Question

a college administrator claims that 78% of college students purchase their books from bookstore. you think this is inaccurate and form a random sample of 46 students at that find that 43 of them purchased their books from the bookstore. test the administrators 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. $h_0$ ? ? blank
c. enter the alternative hypothesis for this test. $h_1$: ? ? blank
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? blank
f. what is the p - value for this test? blank
g. what is the decision based on the given statistics? select an answer

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. $H_1:p
eq0.78$.

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