A presidential candidate's aide estimates that, among all college students, the proportion p who intend to vote in the upcoming election is at least 60%. out of a random sample of 230 college students expressed an intent to vote, can we reject the aide's estimate at the 0.05 level of significance? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified in the table. (if necessary, consult a listof fermulas-) (0) state the null hypothesis H, and the alternative hypothesis Hq- (b) Determine the type of test statistic to use. (Choose one) D-O Oso D20 (e) Find the value of the test statistic. (Round to three or more decimal places.) ? (4) Find the p-value. (Round to three or more decimal places.) (e) Can we reject the aide's estimate that the proportion of college students who intend to vote is at least 60?

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### Educational Website Transcription: Statistical Hypothesis Testing Example

**Topic: Hypothesis Testing for Proportions**

**Example Scenario:**
A presidential candidate’s aide estimates that, among all college students, the proportion \( p \) who intend to vote in the upcoming election is at least 60%. If 129 out of a random sample of 230 college students expressed an intent to vote, can we reject the aide’s estimate at the 0.05 level of significance?

### Steps to Perform a One-Tailed Test

Carry your intermediate computations to three or more decimal places and round your answers as specified below. (If necessary, consult a list of formulas.)

#### (a) State the hypotheses:
- **Null Hypothesis (\( H_0 \))**: \( p = 0.60 \)
- **Alternative Hypothesis (\( H_1 \))**: \( p < 0.60 \)

This represents a one-tailed test where we are testing if the proportion is less than 60%.

#### (b) Determine the type of test statistic to use:
*Choose one from:*

- Z (for large sample size and proportion)
- T (for smaller sample size or unknown population standard deviation)
- Chi-square
- F

For this scenario, we use the **Z-test** since we are dealing with proportions and have a sufficiently large sample size.

#### (c) Find the value of the test statistic:
Substitute the given values into the Z-test formula for proportions. Here, the formula is:

\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \]

Where:
- \(\hat{p}\) = sample proportion = \(\frac{129}{230}\)
- \(p_0\) = hypothesized population proportion = 0.60
- \(n\) = sample size = 230

Calculate the sample proportion:

\[ \hat{p} = \frac{129}{230} = 0.561 \]

Calculate the standard error:

\[ SE = \sqrt{\frac{0.60 \times 0.40}{230}} = 0.0324 \]

Calculate the Z value:

\[ Z = \frac{0.561 - 0.60}{0.0324} = -1.2037 \]

Round off the answer to
Transcribed Image Text:### Educational Website Transcription: Statistical Hypothesis Testing Example **Topic: Hypothesis Testing for Proportions** **Example Scenario:** A presidential candidate’s aide estimates that, among all college students, the proportion \( p \) who intend to vote in the upcoming election is at least 60%. If 129 out of a random sample of 230 college students expressed an intent to vote, can we reject the aide’s estimate at the 0.05 level of significance? ### Steps to Perform a One-Tailed Test Carry your intermediate computations to three or more decimal places and round your answers as specified below. (If necessary, consult a list of formulas.) #### (a) State the hypotheses: - **Null Hypothesis (\( H_0 \))**: \( p = 0.60 \) - **Alternative Hypothesis (\( H_1 \))**: \( p < 0.60 \) This represents a one-tailed test where we are testing if the proportion is less than 60%. #### (b) Determine the type of test statistic to use: *Choose one from:* - Z (for large sample size and proportion) - T (for smaller sample size or unknown population standard deviation) - Chi-square - F For this scenario, we use the **Z-test** since we are dealing with proportions and have a sufficiently large sample size. #### (c) Find the value of the test statistic: Substitute the given values into the Z-test formula for proportions. Here, the formula is: \[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \] Where: - \(\hat{p}\) = sample proportion = \(\frac{129}{230}\) - \(p_0\) = hypothesized population proportion = 0.60 - \(n\) = sample size = 230 Calculate the sample proportion: \[ \hat{p} = \frac{129}{230} = 0.561 \] Calculate the standard error: \[ SE = \sqrt{\frac{0.60 \times 0.40}{230}} = 0.0324 \] Calculate the Z value: \[ Z = \frac{0.561 - 0.60}{0.0324} = -1.2037 \] Round off the answer to
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