Question

The Institute of Education Sciences measures the high school dropout rate as the percentage of 16- through 24-year-olds who are not enrolled in school and have not earned a high school credential. Last year, the high school dropout rate was 8.1%. A polling company recently took a survey of 1,000 people between the ages of 16 and 24 and found that 6.5% of them are high school dropouts. The polling company would like to determine whether the dropout rate has decreased. At a 5% significance level, the decision is to ________.

Answer 1

Answer:

Step-by-step explanation:

Given that:

The population proportion = 8.1% = 0.081

Sample size = 1000

The sample proportion $\hat p$ = 6.5%  = 0.065

The null and the alternative hypothesis for these studies can be computed as:

$\mathbf{H_o : p = 0.081} \\ \\ \mathbf{H_1 : p \neq 0.081}$

Since $H_1$ is not equal to the population proportion; then this is a two-tailed test.

The level of significance is given as 5% = 0.05

The standard error $\sigma_p$ of the sample proportion $\hat p$ can be computed as:

$\sigma_p = \sqrt{\dfrac{p(1-p)}{n} }$

$\sigma_p = \sqrt{\dfrac{0.081(1-0.081)}{1000} }$

$\sigma_p = \sqrt{\dfrac{0.081(0.919)}{1000} }$

$\sigma_p = 0.0086$

The z score test statistic is calculated as:

$z =\dfrac{\hat p - p}{\sigma_p}$

$z =\dfrac{0.065- 0.081}{0.0086}$

z = −1.860

For a two-tailed test, the p-value = $2 \times P(Z < -|z|)$

$p-value = 2 \times P(Z < - 1.86)$

p-value = 2 × 0.0314

p-value = 0.0628

Rejection Criteria: To reject $H_o$ ; if the p-value is less than the level of significance ∝

The decision rule: We failed to reject the null hypothesis since the p-value is greater than the level of significance ∝.

Thus, there is insufficient evidence to conclude that p is not equal to 0.081