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Citation: Week 5 Homework

Question 2 of 10 (1 point)
Question Attempt: 6 of Unlimited

Among college students, the proportion [tex]\( p \)[/tex] who say they're interested in their congressional district's election results has traditionally been [tex]\( 75\% \)[/tex]. After a series of debates on campuses, a political scientist claims that the proportion of college students who say they're interested in their district's election results is more than [tex]\( 75\% \)[/tex]. A poll is commissioned, and 222 out of a random sample of 270 college students say they're interested in their district's election results. Is there enough evidence to support the political scientist's claim at the 0.01 level of significance?

Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.)

(a) State the null hypothesis [tex]\( H_0 \)[/tex] and the alternative hypothesis [tex]\( H_1 \)[/tex].

[tex]\[
\begin{array}{l}
H_0: p=0.75 \\
H_1: p\ \textgreater \ 0.75
\end{array}
\][/tex]

(b) Determine the type of test statistic to use.



Answer :

GinnyAnswer
To solve this problem, we need to perform a one-tailed hypothesis test to determine if there is enough evidence at the 0.01 significance level to support the political scientist's claim that more than 75% of college students are interested in their district's election results.

Let's go through the steps in detail:

### (a) State the null hypothesis [tex]\( H_0 \)[/tex] and the alternative hypothesis [tex]\( H_1 \)[/tex].
The null hypothesis ([tex]\( H_0 \)[/tex]) and the alternative hypothesis ([tex]\( H_1 \)[/tex]) are as follows:

[tex]\[ H_0: p = 0.75 \][/tex]
[tex]\[ H_1: p > 0.75 \][/tex]

Here, [tex]\( p \)[/tex] represents the actual proportion of college students who are interested in their district's election results.

### (b) Determine the type of test statistic to use.
Given that we are dealing with proportions and our sample size is large, we will use a z-test for proportions.

### Step-by-Step Solution:

1. Calculate the sample proportion ([tex]\( \hat{p} \)[/tex]):

[tex]\[ \hat{p} = \frac{222}{270} \approx 0.822 \][/tex]

2. Determine the population proportion under the null hypothesis ([tex]\( p \)[/tex]):

[tex]\[ p = 0.75 \][/tex]

3. Calculate the standard error (SE) of the sample proportion:

The standard error of the sample proportion is given by:

[tex]\[ SE = \sqrt{\frac{p(1 - p)}{n}} \][/tex]

where [tex]\( n \)[/tex] is the sample size.

[tex]\[ SE = \sqrt{\frac{0.75 \times (1 - 0.75)}{270}} \approx 0.026 \][/tex]

4. Calculate the test statistic (z):

The z-score is calculated using the formula:

[tex]\[ z = \frac{\hat{p} - p}{SE} \][/tex]

Plugging in our values:

[tex]\[ z = \frac{0.822 - 0.75}{0.026} \approx 2.741 \][/tex]

5. Find the critical value for the one-tailed test at the 0.01 significance level:

For a one-tailed test at the 0.01 significance level, we look up the critical value in the z-table, which corresponds to 2.326.

6. Comparison and Conclusion:

We compare the calculated test statistic (2.741) to the critical value (2.326). If the test statistic is greater than the critical value, we reject the null hypothesis.

Since [tex]\( 2.741 > 2.326 \)[/tex], we reject the null hypothesis.

### Conclusion:
There is enough evidence at the 0.01 significance level to support the political scientist's claim that more than 75% of college students are interested in their district's election results.

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