a college administrator claims that 78% of college students purchase their books from bookstore. you think…

Explanation:

Step1: Identify the test type

This is a one - proportion z - test as we are testing a claim about a population proportion.

Step2: State the null hypothesis

The administrator claims that 78% of college students purchase books from the bookstore. So, $H_0:p = 0.78$.

Step3: State the alternative hypothesis

We think the claim is inaccurate, so it's a two - tailed test. $H_1:p\neq0.78$.

Step4: Locate the original claim

The original claim is $p = 0.78$, which is in the null hypothesis.

Step5: Calculate the sample proportion

The sample size $n = 46$ and the number of successes $x = 43$. The sample proportion $\hat{p}=\frac{x}{n}=\frac{43}{46}\approx0.9348$.

Step6: Calculate the test statistic

The formula for the one - proportion z - test statistic is $z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}$. Substituting $p = 0.78$, $\hat{p}=0.9348$, and $n = 46$, we get: [ \begin{align*} z&=\frac{0.9348 - 0.78}{\sqrt{\frac{0.78\times(1 - 0.78)}{46}}}\ &=\frac{0.1548}{\sqrt{\frac{0.78\times0.22}{46}}}\ &=\frac{0.1548}{\sqrt{\frac{0.1716}{46}}}\ &=\frac{0.1548}{\sqrt{0.00373}}\ &=\frac{0.1548}{0.0611}\ &\approx2.53 \end{align*} ]

Step7: Calculate the p - value

For a two - tailed test with $z\approx2.53$, the p - value is $2\times(1 - P(Z<2.53))$. From the standard normal table, $P(Z < 2.53)=0.9943$. So the p - value is $2\times(1 - 0.9943)=2\times0.0057 = 0.0114$.

Step8: Make a decision

The level of significance $\alpha=0.10$. Since the p - value $0.0114<0.10$, we reject the null hypothesis.

Answer:

a. One - proportion z - test b. $H_0:p = 0.78$ c. $H_1:p\neq0.78$ d. Null hypothesis e. Approximately $2.53$ f. Approximately $0.0114$ g. Reject the null hypothesis