Answer :
Answer:
Reject H₀.
Step-by-step explanation:
In this case, we need to test whether the average student spent less than the recommended amount of time doing homework in statistics.
The provided data is:
S = {20, 29, 28, 22, 26, 22, 22, 18, 23, 21, 20, 27}
Compute the sample mean:
[tex]\bar x=\frac{1}{n}\sum X=\frac{1}{12}\cdot [20+29+...+27]=23.167[/tex]
The population standard deviation is σ = 7.
The hypothesis for the test is:
H₀: The average student does not spent less than the recommended amount of time doing homework, i.e. μ ≥ 24.
Hₐ: The average student spent less than the recommended amount of time doing homework, i.e. μ < 24.
(A)
Compute the standardized test statistic value as follows:
[tex]z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}[/tex]
[tex]=\frac{23.167-24}{7/\sqrt{12}}\\\\=-0.412[/tex]
Thus, the standardized test statistic value is -0.412.
(B)
The significance level of the test is:
α = 0.07
The critical value of z is:
z₀.₀₇ = -1.476
The rejection region is:
(-∞, -0.1476)
(C)
Compute the p-value as follows:
[tex]p-value=P(Z<-0.412)=0.34[/tex]
*Use a z-table.
Thus, the p-value is 0.34.
(D)
Since, p-value = 0.34 > α = 0.07, the null hypothesis was failed to be rejected at 7% level of significance.
Thus, the correct option is (A).