Chapter 4

I created std error function

std_err <- function(deviation,n){
    standard_error <- deviation/(sqrt(n))
    return(standard_error)
}
std_err(1.75,110)

## [1] 0.166856

std_err(4.15,52)

## [1] 0.5755015

 a <- 7.42
 s <- 1.75
 n <- 110
 error <- qt(0.975,df=n-1)*s/sqrt(n)
 left <- a-error
 right <- a+error
 print(paste(left,":", right))

## [1] "7.08929692545117 : 7.75070307454883"

4.4

A. What is the point estimate for the average height of active individuals? What about the median?

• The point estimate is the same as the mean therefore it is 171.1. The median is listed at 170.3

B. What is the point estimate for the standard deviation of the heights of active individuals? What about the IQR?

• 1 $\sigma$ + 171.1= 180.5 the IQR= 180.5-161.7 =18.8

171.1-9.4

## [1] 161.7

180.5-161.7

## [1] 18.8

C. Is a person who is 1m 80cm (180 cm) tall considered unusually tall? And is a person who is 1m 55cm (155cm) considered unusually short? Explain your reasoning.

• 180 cm falls within 1 standard deviation of the average height. That implies they are taller than about 83% of all people. That qualifies as tall, although I don’t know if it qualifies as unusually tall. 155 cm is about 1.5 $\sigma$ which means 96% of people are taller, that definitely qualifies as unusually short

small <- 171.1-155
tall <- 180-171.1
paste(small/9.4, tall/9.4, sep=" /")

## [1] "1.71276595744681 /0.946808510638298"

D. The mean and the deviation wouldn’t be exactly the same due to randomization which is why we build confidence intervals to determine if a change in the mean is simply do to chance or can be attributed to an external factor.

E. We use the standard error

std_err <- function(deviation,n){
    standard_error <- deviation/(sqrt(n))
    return(standard_error)
}
print(std_err(9.4,507))

## [1] 0.4174687

4.14

A. We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11. + False the average spending of this group is the mean which is 84.71

B. This confidence interval is not valid since the distribution of spending in the sample is right skewed.

• False the sample size is likely large enough to make up for that skew

C. 95% of random samples have a sample mean between $80.31 and $89.11.

• False. As we don’t know where the population parameter is we can’t estimate the CI for all the samples. If the mean is at the end of this interval, than other intervals could be nearly another 2 std errors away

D. We are 95% confident that the average spending of all American adults is between $80.31 and

• True, this is the definition of the CI

E. A 90% confidence interval would be narrower than the 95% confidence interval since we don’t need to be as sure about our estimate.

• True it would be less accurate and narrower

F. In order to decrease the margin of error of a 95% confidence interval to a third of what it is now, we would need to use a sample 3 times larger.

• False with a sample 3 times larger the margin of error would be 2.54 and the current margin of error is 4.4. the math is done below by calculating the estimated std_deviation from what we know already

std_err <- function(deviation,n){
    standard_error <- deviation/(sqrt(n))
    return(standard_error)
}
print(std_err(9.4,436))

## [1] 0.4501784

sample_mean <- (89.11-80.31)/2
sample_mean <-  84.71
std_err_now <- (89.11-80.31)/4
std_err_now <- 2.2

## new std_err must be 3 size
tehnical_std_deviation <- 2.2*sqrt(436)
print(std_err(45.9,436))

## [1] 2.198211

print(std_err(45.9,1308))

## [1] 1.269138

1.27*4

## [1] 5.08

G. The margin of error is 4.4.

• True. The margin of error is defined as $$1z(+-)se$$

4.24

A. Are conditions for inference satisfied?

• Conditions are that the sample is over 30, there is no strong skew, and the sample observations are independent Although our N is met, our distribution seems somewhat rightward skewed. We should be careful and note that in our results considering the low n value

B. Suppose you read online that children first count to 10 successfully when they are 32 months old, on average. Perform a hypothesis test to evaluate if these data provide convincing evidence that the average age at which gifted children fist count to 10 successfully is less than the general average of 32 months. Use a significance level of 0.10.

H~O: the difference between the average time a gifted child and a normal child counts to 10 successfully is 0 H~A: the difference between the average time a gifted child and a normal child counts to 10 successfully is >0

actual_diff <- 32-30.69
z <- 1.645
n <- 36
my_val <- std_err(4.31,36)
my_z <- actual_diff/my_val
my_p <- 1- .9656

paste("Answer: since my pscore:",my_p,"is less than:",.10," We can reject the null hypothesis with a significance level of .10")

## [1] "Answer: since my pscore: 0.0344 is less than: 0.1  We can reject the null hypothesis with a significance level of .10"

C. Interpret the p-value in context of the hypothesis test and the data.

• Contextually 1.5 months at around 30 months old is not practically significant. We don’t even know how accurate our population average is to start and our P value while statistically significant, is barely statistically significant at a 95% CI.

D. Calculate a 90% confidence interval for the average age at which gifted children first count to 10 successfully.

• CI= (1.645 + SE)((1.645 - SE)
  • 29.97-31.4

my_ster <-  std_err(4.31,36)
paste(30.69-my_ster,"-",30.69+my_ster)

## [1] "29.9716666666667 - 31.4083333333333"

5. Do your results from the hypothesis test and the confidence interval agree? Explain

• Yes. The results to the hypothesis test show a statistical significance that means there is a greater than 90% likelihood that our gifted mean is independent. Therefore we would expect our new confidence interval at 90% of the sample, to not encompass the population mean which is 32

4.26

1. Perform a hypothesis test to evaluate if these data provide convincing evidence that the average IQ of mothers of gifted children is di↵erent than the average IQ for the population at large, which is 100. Use a significance level of 0.10.

H~O: the difference between the average iq of mothers of gifted children and the population at large = 0 H~A: the difference between the average iq of mothers of gifted children and the population at large != 0

actual_diff <- 118.2-100
z <- 1.645
n <- 36
my_val <- std_err(6.5,36)
my_z <- actual_diff/my_val
my_p <- .0002*2
#I dont know the p at this z level is but its two tailed so it needs to be muuliplyed by 2 

• We can very confidently reject the null hypothesis as the p value here is extremely small.

2. Calculate a 90% confidence interval for the average IQ of mothers of gifted children.

my_ster <-  std_err(6.5,36)
paste(118.2-my_ster,"-",118.2+my_ster)

## [1] "117.116666666667 - 119.283333333333"

C. Do your results from the hypothesis test and the confidence interval agree? Explain.

• The Confidence interval for our mother of gifted child IQ is nowhere near the mean IQ of the population. Out p value very strongly suggests the same conclusion. The statistical significance and practical significance in this example is large

4.34

• The sampling distribution of the mean, is the accumulation of the means of samples from an overall population. The shape of a distribution approaches normal as N increases. The spread narrows as N increases, and the Center becomes taller.

4.40

1. What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?

diff <- 10500-9000
z_score <- diff/1000

Answer 6.7%

B Describe the distribution of the mean lifespan of 15 light bulbs.

• The 95% confidence interval would contain samples of light bulbs with a mean of 8495-9505.
• This sample would be approaching a normal sample with a mean centered at 9000
• standard error is 258

std_err(1000,15)

## [1] 258.1989

9000+1.96*(258)

## [1] 9505.68

9000-1.96*(258)

## [1] 8494.32

pnorm(5.81)

## [1] 1

3. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?

.067^15

## [1] 2.461059e-18

4. Sketch the two distributions (population and sampling) on the same scale.

library('DATA606') 

normalPlot(mean = 9000, sd = 1000)

normalPlot(mean = 9000, sd = 258)

E. Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?

• Because the sample size is so small, no you couldn’t used a skewed distribution as it would violate the conditions for inference

4.48 Suppose you conduct a hypothesis test based on a sample where the sample size is n = 50, and arrive at a p-value of 0.08. You then refer back to your notes and discover that you made a careless mistake, the sample size should have been n = 500. Will your p-value increase, decrease, or stay the same? Explain.

actual_diff <- 32-30.69
z <- 1.645
n <- 36
my_val <- std_err(10,50)


my_val_2 <- std_err(10,500)

paste(my_val,my_val_2,sep=" turns into ")

## [1] "1.41421356237309 turns into 0.447213595499958"

• The P values will decrease. Increasing the N will decrease the standard error. Which will lead to a higher zscore because the zscore is the (actual_mean- observed_mean) / standard error. A higher zscore implies a lower p value