In 2011, a U.S. Census report determined that $$ 55 \% $$ of college students are working students. A researcher thinks this percentage has changed and surveys 176 college students. The researcher reports that 122 of the 176 are working students. Is there evidence to support the researcher's claim at the $$ 1 \% $$ significance level? a. Determine the null and alternative hypotheses. $$ H_{0}: p= $$ (Select the correct symbol and enter the value.) $$ H_{1}: p $$ ?v Determine the test statistic. Round to two decimal places. $$ z= $$ c. Find the $$ p $$ - value. Round to four decimal places. b - value = d. Are the results statistically significant? Oail to reject the null hypothesis. Reject the null hypothesis. e. Write the conclusion. There is not sufficient evidence to support the claim that the percentage of working college students has changed. There is sufficient evidence to support the claim that the percentage of working college students has changed.

Question

UpStudy AI · Accepted Answer

**a. Hypotheses**

The null hypothesis is that the proportion of working students is still $$ 55\% $$. The alternative hypothesis is that the proportion is not $$ 55\% $$. Thus,

$$
H_{0} : p = 0.55
$$
$$
H_{1} : p \neq 0.55
$$

---

**b. Test Statistic**

Given:
- $$ n = 176 $$
- Number of working students observed, $$ x = 122 $$
- Sample proportion, $$ \hat{p} = \frac{122}{176} \approx 0.6932 $$
- $$ p_0 = 0.55 $$

The standard error is computed as:

$$
SE = \sqrt{\frac{p_0 (1-p_0)}{n}} = \sqrt{\frac{0.55 \times 0.45}{176}} = \sqrt{\frac{0.2475}{176}} \approx \sqrt{0.00140625} \approx 0.0375
$$

The $$ z $$ test statistic is:

$$
z = \frac{\hat{p} - p_0}{SE} = \frac{0.6932 - 0.55}{0.0375} \approx \frac{0.1432}{0.0375} \approx 3.82
$$

---

**c. $$ p $$-value**

This is a two-tailed test. The $$ p $$-value is computed as:

$$
p\text{-value} = 2P(Z > 3.82)
$$

Looking up the standard normal table or using a calculator, the tail probability for $$ z = 3.82 $$ is approximately $$ 0.00007 $$. Therefore,

$$
p\text{-value} \approx 2 \times 0.00007 = 0.00014
$$

Rounded to four decimal places, the $$ p $$-value is approximately:

$$
p\text{-value} \approx 0.0001
$$

---

**d. Statistical Significance**

Since the $$ p $$-value ($$ 0.0001 $$) is less than the significance level ($$ \alpha = 0.01 $$), we reject the null hypothesis.

---

**e. Conclusion**

There is sufficient evidence to support the claim that the percentage of working college students has changed.
**a. Hypotheses** - **Null Hypothesis ($$ H_{0} $$)**: The proportion of working students is $$ 55\% $$ ($$ p = 0.55 $$). - **Alternative Hypothesis ($$ H_{1} $$)**: The proportion of working students is not $$ 55\% $$ ($$ p \neq 0.55 $$). --- **b. Test Statistic** - **Sample Proportion ($$ \hat{p} $$)**: $$ \frac{122}{176} \approx 0.6932 $$ - **Standard Error (SE)**: $$ \sqrt{\frac{0.55 \times 0.45}{176}} \approx 0.0375 $$ - **$$ z $$-Statistic**: $$ \frac{0.6932 - 0.55}{0.0375} \approx 3.82 $$ --- **c. $$ p $$-value** - **$$ p $$-value**: Approximately $$ 0.0001 $$ --- **d. Statistical Significance** - **Conclusion**: Since the $$ p $$-value ($$ 0.0001 $$) is less than the significance level ($$ \alpha = 0.01 $$), we reject the null hypothesis. --- **e. Conclusion** - **Result**: There is sufficient evidence to support the claim that the percentage of working college students has changed.

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Statistics Questions