Question

A past study claims that adults in America spend an average of 17 hours a week on leisure activities. A researcher wanted to test this claim. She took a sample of 10 adults and asked them about the time they spent per week on leisure activities. Their responses (in hours) are as follows:13242137152518224032Assume that the times spent on leisure activities by all adults are normally distributed. At the 5% significance level, can you conclude that the claim of this earlier study is true

Answer 1

The claim is false.

Explanation:

Given the data :

13 24 21 37 15 25 18 22 40 32

The sample mean and standard deviation can ben calculated for the given sample.

Using calculator :

Sample mean, xbar = 24.7

Sample standard deviation, s = 9.04

Sample size, n = 10

The hypothesis :

H0 : μ = 17

H1 : μ ≠ 17

The test statistic :

(xbar - μ) ÷ (s/√(n))

(24.7 - 17) ÷ (9.04/√(10))

7.7 / 2.8586990

Test statistic = 2.694

We can obtain the Pvalue, at α = 0.05 ; df = n-1 = 9

Pvalue = 0.0246

Since Pvalue < α ; we reject the null ; Hence, there is significant evidence to conclude that an adult American does not spend average of 17 hours in leisure