Answer:
Reject <em>H</em>₀.
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:
The population standard deviation is <em>σ</em> = 7.
The hypothesis for the test is:
<em>H</em>₀: The average student does not spent less than the recommended amount of time doing homework, i.e. <em>μ</em> ≥ 24.
<em>Hₐ</em>: The average student spent less than the recommended amount of time doing homework, i.e. <em>μ</em> < 24.
(A)
Compute the standardized test statistic value as follows:
Thus, the standardized test statistic value is -0.412.
(B)
The significance level of the test is:
<em>α</em> = 0.07
The critical value of <em>z</em> is:
<em>z</em>₀.₀₇ = -1.476
The rejection region is:
(-∞, -0.1476)
(C)
Compute the <em>p</em>-value as follows:
*Use a <em>z</em>-table.
Thus, the <em>p</em>-value is 0.34.
(D)
Since, <em>p</em>-value = 0.34 > <em>α</em> = 0.07, the null hypothesis was failed to be rejected at 7% level of significance.
Thus, the correct option is (A).