Question

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).

5.14 4.93 4.07 4.64 5.28 4.67

4.15 4.94 5.13 5.53 5.32 4.60

The mean price of this hamburger in the U.S. in January was $4.61. 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= 2.00

P-value= ? ( three decimal places)

A credit bureau analysis of undergraduate students' credit records found that the average number of credit cards in an undergraduate's wallet was 4.08. It was also reported that in a random sample of 135 undergraduates, the sample mean number of credit cards that the students said they carried was 2.8. The sample standard deviation was not reported, but for purposes of this exercise, suppose that it was 1.1. Is there convincing evidence that the mean number of credit cards that undergraduates report carrying is less than the credit bureau's figure of 4.08? (Use

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	=	.000

Answer 1

Soluiton-1:

Perform t .test in R to get the t and p value.

Rcode:

to get T and p value

Europe_hambg <- c(5.14, 4.93, 4.07 ,4.64 ,5.28 ,4.67,
  
4.15, 4.94, 5.13 ,5.53, 5.32 ,4.60)
t.test(Europe_hambg,mu=4.61,alternative = "greater")

Output:

ne Sample t-test

data: Europe_hambg
t = 1.9533, df = 11, p-value = 0.03835
alternative hypothesis: true mean is greater than 4.61
95 percent confidence interval:
4.630679 Inf
sample estimates:
mean of x
4.866667

t=1.95

p=0.038

Solution-2:

Ho:mu=4.08

Ha:mu<4.08

use ti 83cal to get t and p value

t=-13.52

p=0.000