Question

a college administrator claims that 78% of college students purchase their books from the campus bookstore. you think this is inaccurate and form a random sample of 46 students at that college and find that 43 of them purchased their books from the bookstore. test the administrators claim using a level of significance of 10%.
a. what type of test will be used in this problem? select an answer
b. enter the null hypothesis for this test. $h_0$? ?
c. enter the alternative hypothesis for this test. $h_1$: ? ?
d. is the original claim located in the null or alternative hypothesis? select an answer
e. what is the test statistic for the given statistics?
f. what is the p - value for this test?
g. what is the decision based on the given statistics? select an answer

Explanation:

Step1: Identify test type

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

Step2: State null hypothesis

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

Step3: State alternative hypothesis

We think the claim is inaccurate, so it's a two - tailed test. $H_1:p
eq0.78$.

Step4: Locate original claim

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

Step5: Calculate 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 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 $n = 46$:
\[

$$\begin{align*} z&=\frac{0.9348 - 0.78}{\sqrt{\frac{0.78\times(1 - 0.78)}{46}}}\\ &=\frac{0.1548}{\sqrt{\frac{0.78\times0.22}{46}}}\\ &=\frac{0.1548}{\sqrt{\frac{0.1716}{46}}}\\ &=\frac{0.1548}{\sqrt{0.00373}}\\ &=\frac{0.1548}{0.0611}\\ &\approx2.53 \end{align*}$$

\]

Step7: Calculate 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. $H_1:p
eq0.78$
d. Null hypothesis
e. Approximately 2.53
f. 0.0114
g. Reject the null hypothesis