help ### Hypothesis Testing for the Proportion of Men Under 50 Regularly Visiting Their Family Doctor#### ObjectiveYou are conducting a study to see if the proportion of men under 50 who regularly visit their family doctor is significantly less than 0.13. You use a significance level of $\alpha = 0.01$.#### Hypotheses- **Null Hypothesis ($H_0$)**: $ p = 0.13 $- **Alternative Hypothesis ($H_a$)**: $ p < 0.13 $#### Sample DataYou obtain a sample of size 198 in which there are 23 men (under 50) who regularly visit their family doctor.#### Questions to Addressa) What is the test statistic?b) What is the p-value?c) What is the conclusion (in the context of the problem)?#### Detailed Solution**a) What is the test statistic?**The test statistic for a proportion can be calculated using the formula:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \]Where:- $\hat{p}$ is the sample proportion, calculated as $\hat{p} = \frac{x}{n}$ where $x$ is the number of successes and $n$ is the sample size.- $p_0$ is the assumed population proportion under the null hypothesis.- $n$ is the sample size.Given:- $x = 23$- $n = 198$- $p_0 = 0.13$First, calculate $\hat{p}$:\[ \hat{p} = \frac{23}{198} \approx 0.116 \]Now, calculate the standard error:\[ SE = \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.13 \times 0.87}{198}} \approx 0.024 \]Finally, calculate the test statistic $z$:\[ z = \frac{0.116 - 0.13}{0.024} \approx -0.58 \]**b) What is the p-value?**The p-value corresponds to the probability of obtaining a test

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You are conducting a study to see if the proportion of men under 50 who regularly visit their family doctor is significantly less than 0.13. You use a significance level of a=0.01. Но: р%3D0.13 На: р <0.13 You obtain a sample of size 198 where there are 23 men (under 50) who regularly visit their family doctor. a) What is the test statistic? b) What is the p-value? c) What is the conclusion (in the context of the problem)?

You are conducting a study to see if the proportion of men under 50 who regularly visit their family doctor is significantly less than 0.13. You use a significance level of a=0.01. Но: р%3D0.13 На: р <0.13 You obtain a sample of size 198 where there are 23 men (under 50) who regularly visit their family doctor. a) What is the test statistic? b) What is the p-value? c) What is the conclusion (in the context of the problem)?

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### Hypothesis Testing for the Proportion of Men Under 50 Regularly Visiting Their Family Doctor

#### Objective
You are conducting a study to see if the proportion of men under 50 who regularly visit their family doctor is significantly less than 0.13. You use a significance level of $\alpha = 0.01$.

#### Hypotheses
- **Null Hypothesis ($H_0$)**: $ p = 0.13 $
- **Alternative Hypothesis ($H_a$)**: $ p < 0.13 $

#### Sample Data
You obtain a sample of size 198 in which there are 23 men (under 50) who regularly visit their family doctor.

#### Questions to Address
a) What is the test statistic?
b) What is the p-value?
c) What is the conclusion (in the context of the problem)?

#### Detailed Solution

**a) What is the test statistic?**

The test statistic for a proportion can be calculated using the formula:
\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \]

Where:
- $\hat{p}$ is the sample proportion, calculated as $\hat{p} = \frac{x}{n}$ where $x$ is the number of successes and $n$ is the sample size.
- $p_0$ is the assumed population proportion under the null hypothesis.
- $n$ is the sample size.

Given:
- $x = 23$
- $n = 198$
- $p_0 = 0.13$

First, calculate $\hat{p}$:
\[ \hat{p} = \frac{23}{198} \approx 0.116 \]

Now, calculate the standard error:
\[ SE = \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.13 \times 0.87}{198}} \approx 0.024 \]

Finally, calculate the test statistic $z$:
\[ z = \frac{0.116 - 0.13}{0.024} \approx -0.58 \]

**b) What is the p-value?**

The p-value corresponds to the probability of obtaining a test

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