Set Up (4 functions) to better answer the questions.

I. Function to Reject or Not

We write a function which takes in two arguments (numbers here), runs some computations (basic inequality) on them and prints an output based on the computation result -

Prof. Arvind Sharma Code

myp=function(p, alpha){
  if(p<alpha){print('REJECT Ho')}else{print('FAIL 2 REJECT')}
}

Test our function to make sure it is performing as intended -

myp(p = .01, alpha = .05)  # p is less than alpha
## [1] "REJECT Ho"
myp(p = .1,  alpha =  .05) # p is greater than alpha
## [1] "FAIL 2 REJECT"

Now, lets write a bit more complex function (takes in many arguments) that is designed to shade the standard normal distribution as the default option for a 5% double sided hypothesis test and can be adapted for other purposes too. You can chnage the arguments of mu, sig, pcts, color,…

II. Function for Shading Normal

shadenorm = function(below=NULL, above=NULL, pcts = c(0.025,0.975), mu=0, sig=1, numpts = 500, color = "gray", dens = 40,                    justabove= FALSE, justbelow = FALSE, lines=FALSE,between=NULL,outside=NULL){

    if(is.null(between)){
         below = ifelse(is.null(below), qnorm(pcts[1],mu,sig), below)
         above = ifelse(is.null(above), qnorm(pcts[2],mu,sig), above)
    }
    if(is.null(outside)==FALSE){
         below = min(outside)
         above = max(outside)
    }
  
    lowlim = mu - 4*sig                         # min point plotted on x axis
    uplim  = mu + 4*sig                         # max point plotted on x axis
    x.grid = seq(lowlim,uplim, length= numpts)
    dens.all = dnorm(x.grid,mean=mu, sd = sig)
    
    if(lines==FALSE){
          plot(x.grid, dens.all, type="l", xlab="X", ylab="Density")    # label y and x axis
    }

    if(lines==TRUE){
          lines(x.grid,dens.all)
    }
    
    if(justabove==FALSE){
        x.below    = x.grid[x.grid<below]
        dens.below = dens.all[x.grid<below]
        polygon(c(x.below,rev(x.below)),c(rep(0,length(x.below)),rev(dens.below)),col=color,density=dens)
    }
    if(justbelow==FALSE){
        x.above    = x.grid[x.grid>above]
        dens.above = dens.all[x.grid>above]
        polygon(c(x.above,rev(x.above)),c(rep(0,length(x.above)),rev(dens.above)),col=color,density=dens)
    }
    
    if(is.null(between)==FALSE){
         from = min(between)
         to   = max(between)
         x.between    = x.grid[x.grid>from&x.grid<to]
         dens.between = dens.all[x.grid>from&x.grid<to]
         polygon(c(x.between,rev(x.between)),c(rep(0,length(x.between)),rev(dens.between)),col=color,density=dens)
    }
}

# TEST THE FUCTION
shadenorm(mu = 0, sig = 1, pcts = c(0.025,0.975))

# shadenorm(mu = 20, sig = 6, pcts = c(0.025,0.975))

III. Function for for Shading t

shadet = function(below=NULL, above=NULL, pcts = c(0.025,0.975), df=1, numpts = 500, color = "gray", dens = 40,   justabove= FALSE, justbelow = FALSE, lines=FALSE,between=NULL,outside=NULL){

    if(is.null(between)){
         below = ifelse(is.null(below), qt(pcts[1],df), below)
         above = ifelse(is.null(above), qt(pcts[2],df), above)
    }
    if(is.null(outside)==FALSE){
         below = min(outside)
         above = max(outside)
    }
  
    lowlim = -4
    uplim  = 4
    x.grid = seq(lowlim,uplim, length= numpts)
    dens.all = dt(x.grid,df)
    
    if(lines==FALSE){
          plot(x.grid, dens.all, type="l", xlab="X", ylab="Density")
    }

    if(lines==TRUE){
          lines(x.grid,dens.all)
    }
    
    if(justabove==FALSE){
        x.below    = x.grid[x.grid<below]
        dens.below = dens.all[x.grid<below]
        polygon(c(x.below,rev(x.below)),c(rep(0,length(x.below)),rev(dens.below)),col=color,density=dens)
    }
    if(justbelow==FALSE){
        x.above    = x.grid[x.grid>above]
        dens.above = dens.all[x.grid>above]
        polygon(c(x.above,rev(x.above)),c(rep(0,length(x.above)),rev(dens.above)),col=color,density=dens)
    }
    
    if(is.null(between)==FALSE){
         from = min(between)
         to   = max(between)
         x.between    = x.grid[x.grid>from&x.grid<to]
         dens.between = dens.all[x.grid>from&x.grid<to]
         polygon(c(x.between,rev(x.between)),c(rep(0,length(x.between)),rev(dens.between)),col=color,density=dens)
    }
}

# TEST THE FUCTION
shadet(df = 4, pcts = c(0.025,0.975))     # see the area under the tails are further away from the mean 0..

# shadet(df = 120, pcts = c(0.025,0.975))   # t dist converges to normal when we have high degrees o freedom..

Inference

1.

Using traditional methods, it takes 109 hours to receive a basic driving license. A new license training method using Computer Aided Instruction (CAI) has been proposed. A researcher used the technique with 190 students and observed that they had a mean of 110 hours. Assume the standard deviation is known to be 6. A level of significance of 0.05 will be used to determine if the technique performs differently than the traditional method. Make a decision to reject or fail to reject the null hypothesis. Show all work in R. ### i. Null and Alternative Hypothesis Ho: \(\mu=109\), Ha: \(\mu \neq 109\) ## Set up Hypothesis ## Ho: the technique does not performs differently than the traditional method. ## Ha:the technique performs differently than the traditional method. x<-110 x1<-109 # t value

t<-(110-109)/6 t Z<-pnorm(t)

190 students

n<-190

Standard deviation

sd<-6 ## Degree of freedom n-1. How many independent pieces of information

df<-n-1 df

Calculating the Se Standard Error

se<-sd/sqrt(n) se

Compute

pi = 2

\(pi=2\)

\(\pi=2\) ## Decide Alpha \(\alpha=.05\) alpha<-0.05

An Area Between - Find a positive number z so that the area under the standard normal curve between −z and z is 0.95.

Here’s the solution. If 95% of the area lies between −z and z , then 5% of the area must lie outside of this range. since normal curves are symmetric, half of this amount–2.5%–must lie before −z . Then the area under the curve before z must be:

pvalue<-0.025+0.95 pvalue

# this is my calculation
x<-110
x1<-109
n<-190
sd<-6
df<-n-1
df
## [1] 189
# Z= (value – mean)/ (Standard Deviation)
Ztesttval<-(x-x1)/(sd/sqrt(n))
Ztesttval
## [1] 2.297341
pval<-pnorm(Ztesttval)
pval
## [1] 0.9892004
pvalue<-2*(1-pval)
pvalue
## [1] 0.0215993
alpha<-0.05
alpha
## [1] 0.05
# compute critical value (split alpha since we have double sided hypothesis)
?pnorm
## starting httpd help server ... done
critical_value  <- qnorm(p = .975, 
                          mean = 0,
                          sd = 1
                )
critical_value
## [1] 1.959964
#find Z critical value. Another way to find critical value 
Zcriric<-qnorm(p=.05/2, lower.tail=FALSE)
Zcriric
## [1] 1.959964
tdf<-qt((p=alpha/2),df,lower.tail=FALSE)
tdf
## [1] 1.972595

p value using a t distribution

Finding the p value using a t distribution and similar to using the Z-score. The difference is that you have to specify the number of degrees of freedom.

sd<-6
x1=109
n<-190
x=110
t<-(110-109)/(sd/sqrt(n))
t
## [1] 2.297341
tdis<-2*pt(t,df=n-1)
tdis
## [1] 1.977305

METHOD 1: Test statistic vs Critical value. If the test statistic is larger than the critical value in absolute terms/“more extreme”, reject the null.

Since the test statistic is more extreme than the critical value here, or alternatively the test statistic lies in the rejection region, we will reject the null.

# shades significance level gates 

shadenorm( mu = 109, 
           sig = 6/sqrt(190), 
           pcts = c(0.025,0.975), 
           color = "orange")   

  # mark point estimate from sample
lines(x   = rep(x = 110,10), 
      y   = seq(from  = 0, 
                to = 1,
                length.out=10), 
      col ='purple')

Alpha, the significance level, is the probability that you will make the mistake of rejecting the null hypothesis when in fact it is true. The p-value measures the probability of getting a more extreme value than the one you got from the experiment. If the p-value is greater than alpha, you accept the null hypothesis.

The smaller the p-value, the stronger the evidence that you should reject the null hypothesis.

A p-value less than 0.05 (typically ≤ 0.05) is statistically significant. …

A p-value higher than 0.05 (> 0.05) is not statistically significant and indicates strong evidence for the null hypothesis.

METHOD 2: Compare the p value with alpha (significance level). If the p value is smaller than alpha, reject the null.

# p value associated with the two sided hypothesis test


pval<-pnorm(Ztesttval)
pval
## [1] 0.9892004
pvalue<-2*(1-pval)
pvalue
## [1] 0.0215993
alpha<-0.05
alpha
## [1] 0.05
### Result: A p-value less than 0.05 (typically ≤ 0.05) is statistically significant. the p value is smaller than alpha, reject the null.

*Since the pvalue is less than alpha, we reject the null hypothesis - unlikely to see the sample mean we saw if the null is true. **

We could have used can use our function to give us the same conclusion (which operationalizes the reject/do not reject rule by comparing p value and alpha/level of significance) -

myp(pvalue, alpha)
## [1] "REJECT Ho"
?qnorm
qnorm(p = Ztesttval)
## Warning in qnorm(p = Ztesttval): NaNs produced
## [1] NaN
 # shades p-values here 
shadenorm( mu = 109, 
           sig = 6/sqrt(190), 
           pcts = c(1-pnorm(Ztesttval),pnorm(Ztesttval)), 
           color = "green")   

  # mark point estimate from sample
lines(x   = rep(x = 110,10), 
      y   = seq(from  = 0, 
                to = 1,
                length.out=10), 
      col ='blue')

rnorm(n, # Number of observations to be generated mean = 110, # Integer or vector of means sd = 6) # Integer or vector of standard deviations Simulated p value could have resulted in the same conclusion. NOT REQUIRED/WILL NOT BE TESTED. The idea below is based on the assumption

  # NOT REQUIRED - Simulated p value gives us pretty much the same result...
temp = rnorm(n = 10000,
             mean = 110,
             sd = 6/sqrt(190))   # 100000 obs generated from mean 110 and sd 6/sqrt(190)

p_value_sim <- 2 * length(temp[temp<=109]) / length(temp)   
p_value_sim 
## [1] 0.0186

METHOD 3: Confidence Interval constructed from the sample point estimate contains the hpyothesized values? If not, reject the null.

CI give us the same conclusion, as the hypothesized population parameter is not within the 95% CI constructed from sample. ** confidence interval is an interval that contains the population parameter with probability 1−α . Calculate the mean Calculate the standard error of the mean Find the t-score that corresponds to the confidence level Calculate the margin of error and construct the confidence interval

alpha <-0.05
n<-190
df<-n - 1
tdf<-qt((p=alpha/2),df,lower.tail=FALSE)
tdf
## [1] 1.972595
x  <- 110
Se    <- 6/sqrt(190)   
z     <- qnorm(p = .975)

interval = c( x - z * Se, x + z * Se)

interval
## [1] 109.1469 110.8531
upper<- x + (tdf*Se)
upper
## [1] 110.8586
lower<- x - (tdf*Se)

lower
## [1] 109.1414

##Conclusion : Since 109 falls outside of the interval, we can reject the null hypothesis and assume that the technique performs differently than the traditional method.

2.

Our environment is very sensitive to the amount of ozone in the upper atmosphere.
The level of ozone normally found is 5.3 parts/million (ppm). A researcher believes that the current ozone level is at an insufficient level. The mean of 5 samples is 5.0 ppm with a standard deviation of 1.1. Does the data support the claim at the 0.05 level?
Assume the population distribution is approximately normal.

  1. i. Null and Alternative Hypothesis

    Ho: \(\mu=5\), Ha: \(\mu \neq 5\) ## Set up Hypothesis ## Ho: Data does support the claim. ## Ha: Data does not support the claim. x<-5.3 x1<-5 # t value

t<-(5.3-5)/1.1 t Z<-pnorm(t)

5 Ozone level

n<-5

Standard deviation

sd<-1.1

Degree of freedom n-1. How many independent pieces of information

df<-n-1 df

Calculating te se Standard Error

se<-sd/sqrt(n) se

Compute

pi = 2

\(pi=2\)

\(\pi=2\) ## Decide Alpha \(\alpha=.05\) alpha<-0.05

An Area Between - Find a positive number z so that the area under the standard normal curve between −z and z is 0.95.

Here’s the solution. If 95% of the area lies between −z and z , then 5% of the area must lie outside of this range. since normal curves are symmetric, half of this amount–2.5%–must lie before −z . Then the area under the curve before z must be:

pvalue<-0.025+0.95 pvalue

x<-5.3
x1<-5
n<-5
sd<-1.1
df<-n-1
df
## [1] 4
# Z= (value – mean)/ (Standard Deviation)
Ztesttval2<-(x-x1)/(sd/sqrt(n))
Ztesttval2
## [1] 0.6098367
pval2<-pnorm(Ztesttval2)
pval2
## [1] 0.729015
pvalue<-2*(1-pval2)
pvalue
## [1] 0.54197
alpha<-0.05
alpha
## [1] 0.05
# compute critical value (split alpha since we have double sided hypothesis)
?pnorm
critical_value2  <- qnorm(p = .975, 
                          mean = 0,
                          sd = 1
                )
critical_value2
## [1] 1.959964
#find Z critical value. Another way to find critical value 
Zcriric2<-qnorm(p=.05/2, lower.tail=FALSE)
Zcriric2
## [1] 1.959964
tdf2<-qt((p=alpha/2),df,lower.tail=FALSE)
tdf2
## [1] 2.776445

## p value using a t distribution Finding the p value using a t distribution and similar to using the Z-score. The difference is that you have to specify the number of degrees of freedom.

** This case Z distribution matches with T Distribution

sd<-1.1
x1=5
n<-5
x=5.3
t<-(5.3-5)/(sd/sqrt(n))
t
## [1] 0.6098367
tdis2<-2*pt(t,df=n-1)
tdis2
## [1] 1.425086

METHOD 1: Test statistic vs Critical value. If the test statistic is larger than the critical value in absolute terms/“more extreme”, reject the null.

Since the test statistic is more extreme than the critical value here, or alternatively the test statistic lies in the rejection region, we will reject the null.

# shades significance level gates 

shadenorm( mu = 5, 
           sig = 1.1/sqrt(5), 
           pcts = c(0.025,0.975), 
           color = "orange")   

  # mark point estimate from sample
lines(x   = rep(x = 5.3,10), 
      y   = seq(from  = 0, 
                to = 1,
                length.out=10), 
      col ='purple')

Alpha, the significance level, is the probability that you will make the mistake of rejecting the null hypothesis when in fact it is true. The p-value measures the probability of getting a more extreme value than the one you got from the experiment. If the p-value is greater than alpha, you accept the null hypothesis.

** The smaller the p-value, the stronger the evidence that you should reject the null hypothesis. ** A p-value less than 0.05 (typically ≤ 0.05) is statistically significant. … ** A p-value higher than 0.05 (> 0.05) is not statistically significant and indicates strong evidence for the null hypothesis.

METHOD 2: Compare the p value with alpha (significance level). If the p value is smaller than alpha, reject the null.

# p value associated with the two sided hypothesis test


pval2<-pnorm(Ztesttval)
pval2
## [1] 0.9892004
pvalue<-2*(1-pval)
pvalue
## [1] 0.0215993
alpha<-0.05
alpha
## [1] 0.05
## Result: A p-value less than 0.05 (typically ≤ 0.05) is statistically significant. the p value is smaller than alpha, reject the null.

A pvalue higher than 0.05 (> 0.05) is not statistically significant and indicates strong evidence for the null hypothesis.

We could have used can use our function to give us the same conclusion (which operationalizes the reject/do not reject rule by comparing p value and alpha/level of significance) -

myp(pvalue, alpha)
## [1] "REJECT Ho"
?qnorm
qnorm(p = Ztesttval2)
## [1] 0.2788935
 # shades p-values here 
shadenorm( mu = 5, 
           sig = 1.1/sqrt(5), 
           pcts = c(1-pnorm(Ztesttval2),pnorm(Ztesttval2)), 
           color = "maroon")   

  # mark point estimate from sample
lines(x   = rep(x = 5.3,10), 
      y   = seq(from  = 0, 
                to = 1,
                length.out=10), 
      col ='yellow')

METHOD 3: Confidence Interval constructed from the sample point estimate contains the hpyothesized values? If not, reject the null.

CI give us the same conclusion, as the hypothesized population parameter is not within the 95% CI constructed from sample. ** confidence interval is an interval that contains the population parameter with probability 1−α . Calculate the mean Calculate the standard error of the mean Find the t-score that corresponds to the confidence level Calculate the margin of error and construct the confidence interval

alpha <-0.05
n<-5
df<-n - 1
tdf<-qt((p=alpha/2),df,lower.tail=FALSE)
tdf
## [1] 2.776445
x     <- 5.3
Se    <- 1.1/sqrt(5)   
z     <- qnorm(p = .975)

interval = c( x - z * Se, x + z * Se)

interval
## [1] 4.335825 6.264175
upper<- x + (tdf*Se)
upper
## [1] 6.66583
lower<- x - (tdf*Se)
lower
## [1] 3.93417

With a 95% confidence interval, the level of ozone normally found is between 4.335825 and 6.264175 ppm.

Because 5.3 is within that interval, we cannot reject the null hypothesis. The data does not support the claim.

Q3 # 3. Our environment is very sensitive to the amount of ozone in the upper atmosphere.
The level of ozone normally found is 7.3 parts/million (ppm). A researcher believes that the current ozone level is not at a normal level. The mean of 51 samples is 7.1 ppm with a variance of 0.49. Assume the population is normally distributed. A level of significance of 0.01 will be used. Show all work and hypothesis testing steps.

Ho: The population is normally distributed Ha: The population is not normally distributed ### i. Null and Alternative Hypothesis Ho: \(\mu=5\), Ha: \(\mu \neq 5\) ## Set up Hypothesis ## Ho: The population is normally distributed.Data doessupport the claim. ## Ha: The population is not normally distributed Data does not support the claim. x<-7.3 x1<-7.1 # t value

t<-(7.3-7)/0.49 t Z<-pnorm(t)

mean of 51

n<-51

Standard deviation

sd<-0.49 ## Degree of freedom n-1. How many independent pieces of information

df<-n-1 df

Calculating he SE Standard Error

se<-sd/sqrt(n) se

Compute

pi = 2

\(pi=2\)

\(\pi=2\) ## Decide Alpha \(\alpha=0.01\) alpha<-0.01

An Area Between - Find a positive number z so that the area under the standard normal curve between −z and z is 0.95.

Here’s the solution. If 95% of the area lies between −z and z , then 5% of the area must lie outside of this range. since normal curves are symmetric, half of this amount–2.5%–must lie before −z . Then the area under the curve before z must be:

pvalue<-0.025+0.95 pvalue

x<-7.3
x1<-7.1
n<-51
sd<-0.49
df<-n-1
df
## [1] 50
# Z= (value – mean)/ (Standard Deviation)
Ztesttval<-(x-x1)/(sd/sqrt(n))
Ztesttval
## [1] 2.914869
pval<-pnorm(Ztesttval)
pval
## [1] 0.9982208
pvalue<-2*(1-pval)
pvalue
## [1] 0.003558382
alpha<-0.01
alpha
## [1] 0.01
# compute critical value (split alpha since we have double sided hypothesis)
?pnorm
critical_value  <- qnorm(p = .975, 
                          mean = 0,
                          sd = 1
                )
critical_value
## [1] 1.959964
#find Z critical value. Another way to find critical value 
Zcriric<-qnorm(p=0.01/2, lower.tail=FALSE)
Zcriric
## [1] 2.575829
tdf<-qt((p=alpha/2),df,lower.tail=FALSE)
tdf
## [1] 2.677793

## p value using a t distribution Finding the p value using a t distribution and similar to using the Z-score. The difference is that you have to specify the number of degrees of freedom.

sd<-0.49
x1=7.1
n<-51
x=7.3
t<-(7.3-7.1)/(sd/sqrt(n))
t
## [1] 2.914869
tdis<-2*pt(t,df=n-1)
tdis
## [1] 1.994688

METHOD 1: Test statistic vs Critical value. If the test statistic is larger than the critical value in absolute terms/“more extreme”, reject the null.

Since the test statistic is more extreme than the critical value here, or alternatively the test statistic lies in the rejection region, we will reject the null.

# shades significance level gates 

shadenorm( mu = 7.1, 
           sig = 0.49/sqrt(51), 
           pcts = c(0.025,0.975), 
           color = "orange")   

  # mark point estimate from sample
lines(x   = rep(x = 7.3,10), 
      y   = seq(from  = 0, 
                to = 1,
                length.out=10), 
      col ='purple')

Alpha, the significance level, is the probability that you will make the mistake of rejecting the null hypothesis when in fact it is true. The p-value measures the probability of getting a more extreme value than the one you got from the experiment. If the p-value is greater than alpha, you accept the null hypothesis.

The smaller the p-value, the stronger the evidence that you should reject the null hypothesis.

A p-value less than 0.05 (typically ≤ 0.05) is statistically significant. …

A p-value higher than 0.05 (> 0.05) is not statistically significant and indicates strong evidence for the null hypothesis.

METHOD 2: Compare the p value with alpha (significance level). If the p value is smaller than alpha, reject the null.

# p value associated with the two sided hypothesis test


pval<-pnorm(Ztesttval)
pval
## [1] 0.9982208
pvalue<-2*(1-pval)
pvalue
## [1] 0.003558382
alpha<-0.01
alpha
## [1] 0.01
### Result: A p-value less than 0.05 (typically ≤ 0.05) is statistically significant. the p value is smaller than alpha, reject the null.

*Since the pvalue is less than alpha, we reject the null hypothesis - unlikely to see the sample mean we saw if the null is true. **

We could have used can use our function to give us the same conclusion (which operationalizes the reject/do not reject rule by comparing p value and alpha/level of significance) -

myp(pvalue, alpha)
## [1] "REJECT Ho"
?qnorm
qnorm(p = Ztesttval)
## Warning in qnorm(p = Ztesttval): NaNs produced
## [1] NaN
 # shades p-values here 
shadenorm( mu = 7.1, 
           sig = 0.49/sqrt(51), 
           pcts = c(1-pnorm(Ztesttval),pnorm(Ztesttval)), 
           color = "brown")   

  # mark point estimate from sample
lines(x   = rep(x = 7.3,10), 
      y   = seq(from  = 0, 
                to = 1,
                length.out=10), 
      col ='cyan')

rnorm(n, # Number of observations to be generated mean = 7.1, # Integer or vector of means sd = 0.49) # Integer or vector of standard deviations Simulated p value could have resulted in the same conclusion. NOT REQUIRED/WILL NOT BE TESTED. The idea below is based on the assumption

  # NOT REQUIRED - Simulated p value gives us pretty much the same result...
temp = rnorm(n = 10000,
             mean = 7.3,
             sd = 0.49/sqrt(51))   # 100000 obs generated from mean 7.3 and sd 0.49/sqrt(51)

p_value_sim <- 2 * length(temp[temp<=7.1]) / length(temp)   
p_value_sim 
## [1] 0.003

METHOD 3: Confidence Interval constructed from the sample point estimate contains the hpyothesized values? If not, reject the null.

CI give us the same conclusion, as the hypothesized population parameter is not within the 95% CI constructed from sample. ** confidence interval is an interval that contains the population parameter with probability 1−α . Calculate the mean Calculate the standard error of the mean Find the t-score that corresponds to the confidence level Calculate the margin of error and construct the confidence interval

alpha <-0.01
n<-51
df<-n - 1
tdf<-qt((p=alpha/2),df,lower.tail=FALSE)
tdf
## [1] 2.677793
x  <- 7.1
Se    <- 0.49/sqrt(51)   
z     <- qnorm(p = .975)

interval = c( x - z * Se, x + z * Se)

interval
## [1] 6.96552 7.23448
upper<- x + (tdf*Se)
upper
## [1] 7.283733
lower<- x - (tdf*Se)

lower
## [1] 6.916267

##Conclusion : With a 99% confidence interval, the level of ozone normally found is between 6.96552 and 7.283733 ppm. ## Because 7.3 is outside of that interval, we can reject the null hypothesis and assume that the data supports the researcher’s claim.

4.

A publisher reports that 36% of their readers own a laptop. A marketing executive wants to test the claim that the percentage is actually less than the reported percentage. A random sample of 100 found that 29% of the readers owned a laptop.
Is there sufficient evidence at the 0.02 level to support the executive’s claim? Show all work and hypothesis testing steps. ## State the null hypothesis H0 and alternative hypothesis Ha

\(H_a = \pi <.36\)

\(H_0 = \pi \geq.36\)

Sample count

\(n=100\) n<-100

p<-.36

phat<-.29

q<-1-p

Degree of freedom n-1. How many independent pieces of information

df<-n-1 df tstat=(phat-p)/Se tstat

Decide on the significance level, α.

alpha<-0.02

\(\alpha=.02\)

don’t have sd, so finding Se using Z value

Se = sqrt( p * q / n) Se

Se<- sqrt( p * q / n) Se

ZScore=(phat-p)/Se ZScore

p_value = pnorm(ZTest)

#pnorm(ZTest)

p_value<-pnorm(ZScore) p_value

# don't have sd, so finding Se using Z value
n<-100

p<-.36 

phat<-.29

q<-1-p 

Se = sqrt( p * q / n)
Se
## [1] 0.048
#p value and the zscore, also called the standard score,
tstat=(phat-p)/Se
tstat
## [1] -1.458333
# A p-value falls between 0 and 1 just like the likelihood of an event happening, it gives the probability of a null hypothesis
# Calculatr the CDF for z-score
p_value<-pt(tstat,df=99)
p_value
## [1] 0.07395698
p_value2<-pnorm(tstat)
p_value2
## [1] 0.07237434

p value using a t distribution

Finding the p value using a t distribution and similar to using the Z-score. The difference is that you have to specify the number of degrees of freedom.

t<-(p-phat)/(p/sqrt(n))
t
## [1] 1.944444
tdis<-2*pt(t,df=n-1)
tdis
## [1] 1.94532
# t-score that corresponds to the confidence level
alpha = 0.02
df = n - 1
t.score = qt(p=alpha/2, df,lower.tail=F)
print(t.score)
## [1] 2.364606
# Calculate the margin of error and construct the confidence interval
 me <- t.score * Se
 me
## [1] 0.1135011
 #confidence interval is the mean +/- margin of error
 
  lower <- p - me
  upper <- p + me
  print(c(lower,upper))
## [1] 0.2464989 0.4735011
#Since the test statistic is more extreme than the critical value here, or alternatively the test statistic lies in the rejection region, we will reject the null.**

METHOD 1: Test statistic vs Critical value. If the test statistic is larger than the critical value in absolute terms/“more extreme”, reject the null.

Since the test statistic is more extreme than the critical value here, or alternatively the test statistic lies in the rejection region, we will reject the null.

# shades significance level gates 

shadenorm( mu = .36, 
           sig =Se,
           pcts = c(.05), 
           color = "orange")   

  # mark point estimate from sample
 lines(x   = rep(phat,10), 
      y   = seq(0.20,
                length.out=10), 
      col ='purple')

Alpha, the significance level, is the probability that you will make the mistake of rejecting the null hypothesis when in fact it is true. The p-value measures the probability of getting a more extreme value than the one you got from the experiment. If the p-value is greater than alpha, you accept the null hypothesis.

The smaller the p-value, the stronger the evidence that you should reject the null hypothesis.

A p-value less than 0.05 (typically ≤ 0.05) is statistically significant. …

A p-value higher than 0.05 (> 0.05) is not statistically significant and indicates strong evidence for the null hypothesis.

METHOD 2: Compare the p value with alpha (significance level). If the p value is smaller than alpha, reject the null.

# p value associated with the two sided hypothesis test
p_value2
## [1] 0.07237434
alpha<-0.02
alpha
## [1] 0.02
## A p-value higher than 0.05 (> 0.05) is not statistically significant and indicates strong evidence for the null hypothesis.

We could have used can use our function to give us the same conclusion (which operationalizes the reject/do not reject rule by comparing p value and alpha/level of significance) -

myp(p_value2, alpha)
## [1] "FAIL 2 REJECT"
?qnorm
#qnorm(p = tsat)

 # shades p-values here 
shadenorm( mu = .36, 
           sig = Se, 
           pcts = c(pnorm(tstat),1-pnorm(tstat)), 
           color = "green")   

  # mark point estimate from sample
lines(x   = rep(p,10), 
      y   = seq(from  = 0, 
                to = 1,
                length.out=10), 
      col ='blue')

METHOD 3: Confidence Interval constructed from the sample point estimate contains the hpyothesized values? If not, reject the null.

CI give us the same conclusion, as the hypothesized population parameter is not within the 95% CI constructed from sample. ** confidence interval is an interval that contains the population parameter with probability 1−α . Calculate the mean Calculate the standard error of the mean Find the t-score that corresponds to the confidence level Calculate the margin of error and construct the confidence interval

  # t-score that corresponds to the confidence level
alpha = 0.02
df = n - 1
t.score = qt(p=alpha/2, df,lower.tail=FALSE)
print(t.score)
## [1] 2.364606
# Calculate the margin of error and construct the confidence interval
 me <- t.score * Se
 me
## [1] 0.1135011
 #confidence interval is the mean +/- margin of error
 ##interval
  lower <- p - me
  upper <- p + me
  print(c(lower,upper))
## [1] 0.2464989 0.4735011
## With a 98% confidence interval, the percent of readers that own a laptop is between 24.64989 and 47.16647 percent. 
##  Because 36 is equal to or less than the upper bound, we cannot reject the null hypothesis. The data does not support the claim.

5.

A hospital director is told that 31% of the treated patients are uninsured. The director wants to test the claim that the percentage of uninsured patients is less than the expected percentage. A sample of 380 patients found that 95 were uninsured. Make the decision to reject or fail to reject the null hypothesis at the 0.05 level. Show all work and hypothesis testing steps.

#State the null hypothesis H0 and alternative hypothesis Ha

patients Sample count

n<-380

p<-.31

phat<-95/380 phat

q<-1-p

Degree of freedom n-1. How many independent pieces of information

df<-n-1 df

Decide on the significance level, α.

alpha<-0.05

don’t have sd, so finding Se using Z value

Se = sqrt( p * q / n) Se

Se<- sqrt( p * q / n) Se

ZScore=(phat-p)/Se ZScore

#p_value = pnorm(ZTest) #pnorm(ZTest)

p_value<-pnorm(ZScore) p_value

# don't have sd, so finding Se using Z value
n<-380

p<-.31 

phat<-95/380

q<-1-p 

Se = sqrt( p * q / n)
Se
## [1] 0.0237254
#p value and the zscore, also called the standard score,
tstat=(phat-p)/Se
tstat
## [1] -2.528935
# A p-value falls between 0 and 1 just like the likelihood of an event happening, it gives the probability of a null hypothesis
# Calculatr the CDF for z-score
p_value<-pnorm(tstat)
p_value
## [1] 0.005720462

p value using a t distribution

Finding the p value using a t distribution and similar to using the Z-score. The difference is that you have to specify the number of degrees of freedom.

phat<-95/380
phat
## [1] 0.25
t<-(phat-p)/sqrt(p*q/(n))
t
## [1] -2.528935
# tdis<-pt(tstat,df=n-1)
# tdis

# t-score that corresponds to the confidence level
alpha = 0.05
df = n - 1
t.score = qt(p=alpha, df,lower.tail=TRUE)
print(t.score)
## [1] -1.648884
# Calculate the margin of error and construct the confidence interval
 me <- t.score * Se
 me
## [1] -0.03912044
 #confidence interval is the mean +/- margin of error
 
  lower <- phat + me
  ##upper <- p + me
  print(c(lower,phat))
## [1] 0.2108796 0.2500000
#Because 31 is equal to or less than the upper bound, we cannot reject the null hypothesis.**

METHOD 1: Test statistic vs Critical value. If the test statistic is larger than the critical value in absolute terms/“more extreme”, reject the null.

Since the test statistic is more extreme than the critical value here, or alternatively the test statistic lies in the rejection region, we will reject the null.

# shades significance level gates 

shadenorm( mu = .31, 
           sig =Se,
           pcts = c(.05), 
           color = "orange")   

  # mark point estimate from sample
 lines(x   = rep(phat,10), 
      y   = seq(0.20,
                length.out=10), 
      col ='red')

Alpha, the significance level, is the probability that you will make the mistake of rejecting the null hypothesis when in fact it is true. The p-value measures the probability of getting a more extreme value than the one you got from the experiment. If the p-value is greater than alpha, you accept the null hypothesis.

The smaller the p-value, the stronger the evidence that you should reject the null hypothesis.

A p-value less than 0.05 (typically ≤ 0.05) is statistically significant. …

A p-value higher than 0.05 (> 0.05) is not statistically significant and indicates strong evidence for the null hypothesis.

METHOD 2: Compare the p value with alpha (significance level). If the p value is smaller than alpha, reject the null.

# p value associated with the two sided hypothesis test



pvalue<-2*(1-p_value)
pvalue
## [1] 1.988559
alpha<-0.05
alpha
## [1] 0.05
## A p-value higher than 0.05 (> 0.05) is not statistically significant and indicates strong evidence for the null hypothesis.

We could have used can use our function to give us the same conclusion (which operationalizes the reject/do not reject rule by comparing p value and alpha/level of significance) -

myp(pvalue, alpha)
## [1] "FAIL 2 REJECT"
?qnorm
qnorm(p = tstat)
## Warning in qnorm(p = tstat): NaNs produced
## [1] NaN
 # shades p-values here 
shadenorm( mu = .31, 
           sig = Se, 
           pcts = c(1-pnorm(tstat),pnorm(tstat)), 
           color = "green")   

  # mark point estimate from sample
lines(x   = rep(p,10), 
      y   = seq(0.20,
                length.out=10), 
      col ='blue')

METHOD 3: Confidence Interval constructed from the sample point estimate contains the hpyothesized values? If not, reject the null.

CI give us the same conclusion, as the hypothesized population parameter is not within the 95% CI constructed from sample. ** confidence interval is an interval that contains the population parameter with probability 1−α . Calculate the mean Calculate the standard error of the mean Find the t-score that corresponds to the confidence level Calculate the margin of error and construct the confidence interval

  # t-score that corresponds to the confidence level
alpha = 0.05
df = n - 1
t.score = qt(p=alpha/2, df,lower.tail=F)
print(t.score)
## [1] 1.966243
# Calculate the margin of error and construct the confidence interval
 me <- t.score * Se
 me
## [1] 0.04664991
 #confidence interval is the mean +/- margin of error
 ##interval
  lower <- p - me
  upper <- p + me
  print(c(lower,upper))
## [1] 0.2633501 0.3566499
## With a 95% confidence interval, the percent of readers that own a laptop is between 24.64989 and 47.16647 percent. 
##  Because 31 is equal to or less than the upper bound, we cannot reject the null hypothesis. The data does not support the claim.

## confidence interval with qnorm
  
ZCI<-qnorm(1-.05)/2
ZCI
## [1] 0.8224268
CI<-ZCI*Se
CI
## [1] 0.01951241
upperCI<-phat+CI
upperCI
## [1] 0.2695124
lowerCI<-phat-CI
lowerCI
## [1] 0.2304876
print(c(lowerCI,upperCI))
## [1] 0.2304876 0.2695124
## With a 95% confidence interval, the percent of uninsured patients is between 93.04876 and 96.95124 percent.

we cannot reject the null hypothesis. The data does not support the claim.

6.

Find the minimum sample size needed to be 99% confident that the sample’s variance is within 1% of the population’s variance.

7.

A standardized test is given to a sixth-grade class. Historically the mean score has been 112 with a standard deviation of 24. The superintendent believes that the standard deviation of performance may have recently decreased. She randomly sampled 22 students and found a mean of 102 with a standard deviation of 15.4387. Is there evidence that the standard deviation has decreased at the 𝛼𝛼 = 0.1 level? Show all work and hypothesis testing steps.

#Ho: sigma>=24, Ha: sigma<=24

x<-112

sd1<-24

sigma=24

sigma_2=sigma^2

n<-22

x1<-102

s<-15.4387

s_2=s^2

alpha<-0.1

#chi # df<-n-1 # chi=df*s_2/sigma_2

shadechi(df = n-1, pcts = c(.1))

lines(rep(chi,10), seq(0,20,length.out=10),col=‘red’)

?pchisq # distribution function for the chi-squared distribution with df degrees of freedom

p_value <- pchisq(q = chi, df = df, lower.tail = T) ## [1] 0.008549436

p_value

myp(p_value,alpha) ## [1] “REJECT Ho”

meandiff<-x-x1

meandiff

SE<-sqrt((sd12/n)+(sd22/n))

SE

statistic<-(112-102)/SE

statistic

sdf<-sd1/sd2

sdf

x<-112
sd1<-24
n<-22
x1<-102
sd2<-15.4387
alpha<-0.1
df<-n-1

meandiff<-x-x1
meandiff
## [1] 10
SE<-sqrt((sd1^2/n)+(sd2^2/n))
SE
## [1] 6.084083
statistic<-(112-102)/SE
statistic
## [1] 1.643633
sdf<-sd1/sd2
sdf
## [1] 1.554535
SE<-sqrt((sd1^2/n)+(sd2^2/n))
SE
## [1] 6.084083
statistic<-(112-102)/SE
statistic
## [1] 1.643633
# Z= (value – mean)/ (Standard Deviation)
ZtestFirstval<-(x-x1)/(sd1/sqrt(n))
ZtestFirstval
## [1] 1.95434
Ztestsecondval<-(x-x1)/(sd2/sqrt(n))
Ztestsecondval
## [1] 3.03809
pvalFirst<-pnorm(ZtestFirstval)
pvalFirst
## [1] 0.9746695
pvalsecond<-pnorm(Ztestsecondval)
pvalsecond
## [1] 0.9988096
pvalueFirst <-2*(1-pvalFirst)
pvalueFirst
## [1] 0.05066103
pvalueSecond <-2*(1-pvalsecond)
pvalueSecond
## [1] 0.00238083
PValAll<-2*pnorm((statistic),lower.tail=FALSE)
PValAll
## [1] 0.100252
Pvalless<-pnorm(statistic, lower.tail = TRUE)
Pvalless
## [1] 0.949874
Pvalmore<-pnorm(statistic, lower.tail = FALSE)
Pvalmore
## [1] 0.05012601
alpha<-0.05
alpha
## [1] 0.05
LowerCL <- (112 - 102 - SE * qnorm(1 - alpha / 2))
LowerCL
## [1] -1.924584
UpperCL <- (112 - 102 + SE * qnorm(1 - alpha / 2))
UpperCL
## [1] 21.92458
## With a 99% confidence interval, the standard deviation is between -1.924584 and 21.92458 
## #  Because 24 is within that interval, we cannot reject the null hypothesis that SD has not decreased below 24.