Question

7. Test the claim that the population mean time a person spends grocery shopping per week is more than 2 hours. Use the sample data below, and use a 5% significance level. (Since this is a small sample size, assume that the population is approximately normally distributed.) You will have to use your calculator to find the sample mean x and the sample standard deviation S x = 2 hours, 3 hours, 2 hours, 1.5 hours, 4 hours, 1.5 hours Show All Steps including the null hypothesis, the p value, whether you reject the null hypothesis or not, and write a brief conclusion.

Answer 1

Solution:
Given in the question
The claim is that the population mean time a person spends grocery shopping per week is more than 2 hours so Null and alternative hypothesis can be written as
Null hypothesis H0: $\mu$ = 2
Alternate hypothesis Ha: $\mu$ >2
Sample data
2,3,2,1.5,4,1.5
Sample mean Xbar = (2+3+2+1.5+4+1.5)/6 = 14/6 = 2.33
Sample standard deviation S can be calculated as
S= sqrt((Xi-mean)^2)/(N-1)) = sqrt((2-2.33)^2 + (3-2.33)^2 + (2-2.33)^2 + (1.5-2.33)^2 + (4-2.33)^2 + (1.5-2.33)^2)/5) = sqrt(4.83/5) = 0.983
Here we will use t test as sample size is less and population standard deviation is unknown, so test stat value can be calculated as
Test stat = (Xbar - $\mu$ )/S/sqrt(n) = (2.33-2)/0.983/sqrt(6) = 0.83
As this test is right tailed test, so degree of freedom = 6-1 = 5 so p-value = 0.2221
At alpha = 0.05, we are failed to reject the null hypothesis as p-value is more than alpha value i.e. (0.2221>0.05). So we dont have significant evidence to support the claim that the population mean time a person spends grocery shopping per week is more than 2 hours.