Question

Suppose that Michael reads somewhere on the Internet that, on average, middle-aged people between 45 and 65 years old visit 8 different websites per day on weekend days. Michael knows that he cannot trust everything he reads on the Internet, so he decides to conduct his own study. He believes that the middle-aged people in his community visit more than 8 websites per day. He surveys a random sample of 14 middle-aged people in his community and asks them to look at their browsing history and count the number of different websites they visited last Saturday. The following stem and leaf plot shows the number of websites visited last Saturday by each person in the survey. The stem unit is tens and the leaf unit is ones. 0 3 0 6 1 0 1 7 4 6 1 8 7 2 9 8 3 4 Leaf unit = 1 website Michael assumes that the number of websites middle-aged people visit on the weekends is normally distributed, so he decides to perform a t-test with a significance level of a = 0.05 to test the null hypothesis, H:n = 8 against the alternative hypothesis, H M > 8, where u is the population mean. If the requirements for a t-test have not been met, only answer the final question. Otherwise, answer all five of the following questions. First, compute the mean, X, of Michael's sample. Report your answer with two decimals of precision. websites Then, compute the sample standard deviation, s. Report your answer with two decimals of precision. websites

Answer 1

Solution:

Perform in R studio.

Use mean function to get xbar,sd function to get s and t.test function to get t and p value.

websites <- c(3,4,6,6,7,8,10,11,12,13,14,17,18,19)
websites
mean(websites)
sd(websites)
t.test(websites,mu=8,alternative = 'greater')

Untitled7* * Console Terminal Jobs RStudio le Edit Code View Llots Session Build Debug Profile Tools Help - OR . O Go to file/function - Addins Untitled3* X Untitled5* * Untitled8** Untitled Untitled4** Source on Save a . E websites <- C(3,4,6,6,7,8,10,11,12,13,14,17,18,19) 2 websites 3 mean websites 4 sd (websites t.test(websites, mu=8, alternative - 'greater') + Run + - Source - L 4 1 IV II IL D 14 17 10 1 > mean(websites) [1] 10.57143 > sd (websites) [1] 5.184508 > t.test(websites, mu=8, alternative = 'greater') One Sample t-test data: websites t = 1.8558, df = 13, p-value = 0.04314 alternative hypothesis: true mean is greater than 8 95 percent confidence interval: 8.117591 Inf sample estimates: mean of x 10. 57143

xbar=10.57

s=5.18

t=1.856

p=0.043

Michael should make the decision to reject the null hypothesis.The conclusion is that the there is sufficient evidence that the mean number of websites visited by middle aged people on the weekends is greater than 8.