Chapter 5 - Foundations for Inference

Heights of adults

imgage <- "C:/Users/jpsim/Documents/Stat & Probability for Data/heights.png"
include_graphics(imgage)

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

Solution:

head(bdims$hgt)

## [1] 174.0 175.3 193.5 186.5 187.2 181.5

meanhgt <- mean(bdims$hgt)
meanhgt

## [1] 171.1438

medianhgt <- median((bdims$hgt))
medianhgt

## [1] 170.3

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

Solution:

sdhgt <- sd(bdims$hgt)

sdhgt

## [1] 9.407205

IQR(bdims$hgt, na.rm = T)

## [1] 14

Q3 <- quantile(bdims$hgt, 0.75) 
Q1 <- quantile(bdims$hgt, 0.25)  
print (Q3 - Q1)

## 75% 
##  14

3. 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.

Solution:

twoSDpos <- meanhgt + 2*sdhgt 

twoSDneg <- meanhgt - 2*sdhgt 

twoSDpos

## [1] 189.9582

twoSDneg

## [1] 152.3294

twoSDpos1 <- meanhgt + 1*sdhgt 

twoSDneg1 <- meanhgt - 1*sdhgt 


twoSDpos1

## [1] 180.551

twoSDneg1

## [1] 161.7366

4. The researchers take another random sample of physically active individuals. Would you expect the mean and the standard deviation of this new sample to be the ones given above? Explain your reasoning.

Solution:

hist(bdims$hgt, probability = TRUE)
x <- 140:200
y <- dnorm(x = x, mean = meanhgt, sd = sdhgt)
lines(x = x, y = y, col = "blue")

5. The sample means obtained are point estimates for the mean height of all active individuals, if the sample of individuals is equivalent to a simple random sample. What measure do we use to quantify the variability of such an estimate (Hint: recall that \(SD_x = \frac{\sigma}{\sqrt{n}}\) )? Compute this quantity using the data from the original sample under the condition that the data are a simple random sample.

Solution:

sd_x <- sdhgt/sqrt(nrow(bdims))

sd_x

## [1] 0.4177887

Thanksgiving Spending

imgage <- "C:/Users/jpsim/Documents/Stat & Probability for Data/thanks.png"
include_graphics(imgage)

Solutions: a. We are 95% confident that the average spending of these 436 American adults is between $80.31 and $89.11.

Solution:

FALSE, Inference is measured on the population parameter. The CI should not be a representation of the sample but the population.

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

Solution:

False, the data is only slightly skewed.

3. 95% of random samples have a sample mean between $80.31 and $89.11. Solution: False - CI for the mean of a sample is not representative of other sample means.

d.We are 95% confident that the average spending of all American adults is between $80.31 and $89.11.

Solution:

TRUE, the population parameter is estimated by the Point Estimate and the CI, respectively.

5. A 90% confidence interval would be narrower than the 95% confidence interval since we do not need to be as sure about our estimate.

Solution:

TRUE, there is a small percentage of CI, the interval becomes narrow

6. 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.

Solution: False, the sample size should 9 times larger

7. The margin of error is 4.4.

Solution: TRUE, the margin of error is half the CI

Gifted Childern Part I

imgage <- "C:/Users/jpsim/Documents/Stat & Probability for Data/giftp1.png"
include_graphics(imgage)

1. Are conditions for inference satisfied?

Solution: Yes the sampling was random, the sample size is > 30 and the distribution is nearly normal without skewness.

Solution:

H0 : \(\mu\) = 32

HA : \(\mu\) < 32

Calculating the SE:

given std. Dev. (sd) = 4.31

given n = 36

SE = sd/sqrt(n) = 4.31/sqrt(36) = 0.7183

To calculate Z-score:

\(Z_{30.69}\) = (30.69-32)/0.7183 = -1.82

P(Z<-1.82) = 0.034

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

Solution: \(\alpha\) = = 0.1 but since the calculated p-value < 0.1 whch is 0.034, we reject the null hypothesis of Ho

therefore, we can with 90% confidence believe there’s evidence that gifted kids start reading before 32

months old.

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

Solution:

CIupper=30.69+(1.645*(0.7183))
CIupper

## [1] 31.8716

CIlower=30.69−(1.645*(0.7183))
CIlower

## [1] 29.5084

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

Solution: Yes, interval doesn’t contain any 32 months old

Gifted Childern Part II

imgage <- "C:/Users/jpsim/Documents/Stat & Probability for Data/giftp2.png"
include_graphics(imgage)

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

Solution:

H0 : \(\mu\) = 100

HA : \(\mu\) != 100

Calculating the SE:

given std. Dev. (sd) = 6.5

given n = 36

SE = sd/sqrt(n) = 6.5/sqrt(36) = 1.083

To calculate Z-score:

\(Z_{118.2}\)= (118.2-100)/1.083 = 16.80

P(Z!=16.80) = 0

Mean = 118.2
n = 36
SD = 6.5
SE = SD/sqrt(n)

z_score = (Mean -100)/SE
pnorm(z_score)

## [1] 1

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

Solution:

CIupper =118.2+(1.645*(1.083))
CIupper

## [1] 119.9815

CIlower=118.2−(1.645*(1.083))
CIlower

## [1] 116.4185

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

Solution: \(\alpha\) = 0.1, based on calculated p-value of 0.0 , we reject the null hypothesis and beleive that there

are enough evidience at the 90% confidence level the IQ is Not 100. The CLT do not contain 100 IQ.

CLT

Define the term “sampling distribution” of the mean, and describe how the shape, center, and spread of the sampling distribution of the mean change as sample size increases.

Solution: Sampling distribution means taking n samples from the population and measuring the means of these samples These samples’ means which in itself has a distribution; therefore the term sampling distribution. The shape, center and spread of the mean changes as sample size increase. The greater the sample size, the more “bell-shape” the distribution becomes or in other words, the shape becomes more normal, the center approaches the true population mean, and the spread decreases.

CFLB’s

1. A manufacturer of compact fluorescent light bulbs advertises that the distribution of the lifespans of these light bulbs is nearly normal with a mean of 9,000 hours and a standard deviation of 1,000 hours.

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

Solution:

\(Z_{10500}\) = (10500-9000)/1000 = 1.5

P(X>10500) = 0.0668

mean = 9000
sd = 1000
z_score = (10500 - 9000)/sd
prob = 1-pnorm(z_score)
prob

## [1] 0.0668072

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

SE_sample = 1000/sqrt(15)
SE_sample

## [1] 258.1989

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

z_score = (10500 - 9000)/258.2
probability = 1 - pnorm(z_score)
probability

## [1] 3.13392e-09

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

s <- seq(5000,13000,0.01)
plot(s, dnorm(s,9000, 1000), type="l", ylim = c(0,0.002),)
lines(s, dnorm(s,9000, 258.1989), col="blue")

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

the probabilities with a skewed distribution cannot be estimated.

Same Observation, Different Saample Size

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.

Solution:

The p-value will decrease with a larger sample size as the spead of the distribution will narrow causing the standard deviation to decrease with increase in n.