Questions: A government official wanted to know if more than 90% of the residents in a region had an income that was less than 100 thousand dollars. The official randomly sampled 500 residents and found that 463 of them had an income that was less than 100 thousand dollars. The official conducts a one-proportion hypothesis test at the 5% significance level, to test whether the true proportion of residents in the region that have an income that is less than 100 thousand dollars is greater than 90%. (a) H0: p ≤ 0.9 ; Ha: p>0.9, which is a right-tailed test. (b) Use Excel to test whether the true proportion of residents in the region that have an income that is less than 100 thousand dollars is greater than 90%. Identify the test statistic, z, and p-value from the Excel output, rounding to three decimal places.

A government official wanted to know if more than 90% of the residents in a region had an income that was less than 100 thousand dollars. The official randomly sampled 500 residents and found that 463 of them had an income that was less than 100 thousand dollars.

The official conducts a one-proportion hypothesis test at the 5% significance level, to test whether the true proportion of residents in the region that have an income that is less than 100 thousand dollars is greater than 90%.
(a) H0: p ≤ 0.9 ; Ha: p>0.9, which is a right-tailed test.
(b) Use Excel to test whether the true proportion of residents in the region that have an income that is less than 100 thousand dollars is greater than 90%. Identify the test statistic, z, and p-value from the Excel output, rounding to three decimal places.

Transcript text: A government official wanted to know if more than $90 \%$ of the residents in a region had an income that was less than 100 thousand dollars. The official randomly sampled 500 residents and found that 463 of them had an income that was less than 100 thousand dollars. The official conducts a one-proportion hypothesis test at the $5 \%$ significance level, to test whether the true proportion of residents in the region that have an income that is less than 100 thousand dollars is greater than $90 \%$. (a) $H_{0}: p \leq 0.9 ; H_{a}: p>0.9$, which is a right-tailed test. (b) Use Excel to test whether the true proportion of residents in the region that have an income that is less than 100 thousand dollars is greater than $90 \%$. Identify the test statistic, $z$, and pvalue from the Excel output, rounding to three decimal places. Provide your answer below:

Solution

Solution Steps

Step 1: Hypothesis Formulation

We are conducting a one-proportion hypothesis test to determine if more than $90\%$ of the residents in a region have an income less than $100$ thousand dollars. The hypotheses are formulated as follows:

Null Hypothesis: $H_0: p \leq 0.9$
Alternative Hypothesis: $H_a: p > 0.9$

Step 2: Test Statistic Calculation

The test statistic $Z$ is calculated using the formula:

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

Where:

$\hat{p} = \frac{463}{500} = 0.926$
$p_0 = 0.9$
$n = 500$

Substituting the values, we find:

\[ Z = \frac{0.926 - 0.9}{\sqrt{\frac{0.9(1 - 0.9)}{500}}} = 1.938 \]

Step 3: P-value Calculation

The p-value associated with the test statistic $Z = 1.938$ is calculated to be:

\[ \text{P-value} = 0.026 \]

Step 4: Decision Rule

At a significance level of $\alpha = 0.05$:

If the p-value is less than $\alpha$, we reject the null hypothesis.

Since $0.026 < 0.05$, we reject the null hypothesis.

Final Answer

The test statistic is $Z = 1.938$ and the p-value is $0.026$. Therefore, we conclude that there is sufficient evidence to support the claim that more than $90\%$ of the residents have an income less than $100$ thousand dollars.

$\boxed{Z = 1.938, \text{ P-value} = 0.026}$

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