Question

only $mu=15$, $sigma=9$, hours per week preparing for classes by reading, doing homework, studying, etc. a state university develops a program that is designed to increase student motivation to study. a sample of $n = 36$ students completes the program and later reports that they spend $m = 18$ hours per week studying. the university would like to test whether the program increased time spent preparing for class.
a. assuming a two - tailed test, state the null and alternative hypotheses in a sentence that includes the two variables being examined.
b. using the standard four - step procedure, conduct a two - tailed hypothesis test with $alpha=.05$ to evaluate the effect of the program.

Explanation:

Step1: State the null and alternative hypotheses

The null hypothesis \(H_0\) is that the program has no effect, so \(\mu = 15\). The alternative hypothesis \(H_1\) is that the program has an effect, so \(\mu
eq15\).

Step2: Calculate the standard error

The formula for the standard error \(\sigma_M=\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 9\) and \(n = 36\), we have \(\sigma_M=\frac{9}{\sqrt{36}}=\frac{9}{6} = 1.5\).

Step3: Calculate the z - statistic

The formula for the z - statistic is \(z=\frac{M-\mu}{\sigma_M}\). Substituting \(M = 18\), \(\mu=15\), and \(\sigma_M = 1.5\), we get \(z=\frac{18 - 15}{1.5}=\frac{3}{1.5}=2\).

Step4: Determine the critical z - values and make a decision

For a two - tailed test with \(\alpha=0.05\), the critical z - values are \(z=\pm1.96\). Since \(z = 2>1.96\), we reject the null hypothesis.

Answer:

a. \(H_0:\mu = 15\), \(H_1:\mu
eq15\)
b. Reject \(H_0\) (the program has a significant effect as \(z = 2>1.96\))