Question

Consumer Reports tested 14 brands of vanilla yogurt and found the following numbers of calories per serving:

160 200 220 230 120 180 140 130 170 190 80 120 100 170

a. Use the R command qqnorm(x) where x is the data set to obtain the normal probability plot of the data set. Is it reasonable to assume the data set is normally distributed? Why or why not?

b. Regardless of the answer in part (a), assume the data set is normally distributed. Create a 99% confidence interval for the mean calorie content of vanilla yogurt.

c. A diet guide claims that you will get 120 calories from a serving of vanilla yogurt. What does the confidence interval in part (b) tell you about this claim? Why?

d. Use a computer software to answer the following question: Assuming the data set is normally distributed. Is there evidence that the mean is different from the diet guide’s claim, which is stated in part (c)?

Answer 1

a. Use the R command qqnorm(x) where x is the data set to obtain the normal probability plot of the data set. Is it reasonable to assume the data set is normally distributed? Why or why not?

Ans -

> qqnorm(c(160, 200, 220, 230, 120, 180, 140, 130, 170, 190, 80, 120, 100, 170))
>

QQ Plot -

YES. It is reasonable to assume that data is Normally distributed because of theabove plot. WHere all the quantiles are situated on the line with minimal distance.

b) a 99% confidence interval for the mean calorie content of vanilla yogurt

> mean(c(160, 200, 220, 230, 120, 180, 140, 130, 170, 190, 80, 120, 100, 170))
[1] 157.8571

c(160, 200, 220, 230, 120, 180, 140, 130, 170, 190, 80, 120, 100, 170))
[1] 44.75206
>

Calculation

M = 158
t = 2.58
s_M = √(44.8²/14) = 11.97

μ = M ± Z(s_M)
μ = 158 ± 2.58*11.97
μ = 158 ± 30.84

M = 158, 99% CI [127.16, 188.84].

C) From the above part

You can be 99% confident that the population mean (μ) falls between 127.16 and 188.84 and A diet guide claims that you will get 120 calories from a serving of vanilla yogurt.

Hence the claim is NOT TRUE because 120 calories is not lying within the 99% confidence interval.

D) Test of Hypothesis

From the t-table with degrees of freedom = 13 and α=0.01

The critical value is 3.012

Conclusion - The calculated t value is smaller than critical value (0.0707<3.012), so the mean of data set is not significantly different from μ0=120.