Question

1) A magazine collects data each year on the price of a hamburger in a certain...

1) A magazine collects data each year on the price of a hamburger in a certain fast food restaurant in various countries around the world. The price of this hamburger for a sample of restaurants in Europe in January resulted in the following hamburger prices (after conversion to U.S. dollars).

5.19 4.95 4.08 4.69 5.22 4.68
4.13 4.99 5.15 5.55 5.34 4.60

The mean price of this hamburger in the U.S. in January was $4.63. For purposes of this exercise, assume it is reasonable to regard the sample as representative of these European restaurants. Does the sample provide convincing evidence that the mean January price of this hamburger in Europe is greater than the reported U.S. price? Test the relevant hypotheses using

α = 0.05.

(Use a statistical computer package to calculate the P-value. Round your test statistic to two decimal places and your P-value to three decimal places.)

t =
P-value =

b) Five students visiting the student health center for a free dental examination during National Dental Hygiene Month were asked how many months had passed since their last visit to a dentist. Their responses were as follows.

6     16     10     24     29    

Assuming that these five students can be considered a random sample of all students participating in the free checkup program, construct a 95% confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program. (Give the answer to two decimal places.)
( , )

Homework Answers

Answer #1

a)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 4.63
Alternative Hypothesis, Ha: μ > 4.63

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (4.8808 - 4.63)/(0.4592/sqrt(12))
t = 1.89

P-value Approach
P-value = 0.043

b)

sample mean, xbar = 17
sample standard deviation, s = 9.5394
sample size, n = 5
degrees of freedom, df = n - 1 = 4

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.776


ME = tc * s/sqrt(n)
ME = 2.776 * 9.5394/sqrt(5)
ME = 11.843

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (17 - 2.776 * 9.5394/sqrt(5) , 17 + 2.776 * 9.5394/sqrt(5))
CI = (5.16 , 28.84)

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