According to a large poll in a previous year, about $80 \%$ of homes in a certain county had access to high-speed internet. Market researchers wondered if that proportion had changed, so they took a random sample of 64 homes from that county and found that 48 of them had access to highspeed internet.

They want to use this sample data to test $H_{0}: p=0.8$ versus $H_{\mathrm{a}}: p \neq 0.8$, where $p$ is the proportion of homes in this county with high-speed internet access.

Assuming that the conditions for inference have been met, calculate the test statistic for their significance test.
You may round to two decimal places.
$$
z=
$$

Question

According to a large poll in a previous year, about $80 \%$ of homes in a certain county had access to high-speed internet. Market researchers wondered if that proportion had changed, so they took a random sample of 64 homes from that county and found that 48 of them had access to highspeed internet.

They want to use this sample data to test $H_{0}: p=0.8$ versus $H_{\mathrm{a}}: p \neq 0.8$, where $p$ is the proportion of homes in this county with high-speed internet access.

Assuming that the conditions for inference have been met, calculate the test statistic for their significance test.
You may round to two decimal places.
$$
z=
$$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Calculate the sample proportion $\hat{p}$ by dividing the number of homes with high-speed internet by the total number of homes in the sample. ‖ ‖ $\hat{p} = \frac{48}{64}$ ‖ ‖ step 2 ⋮ Compute the standard error (SE) using the null hypothesis proportion $p_0$ and the sample size $n$. ‖ ‖ $SE = \sqrt{\frac{p_0(1 - p_0)}{n}}$ where $p_0 = 0.8$ and $n = 64$. ‖ ‖ step 3 ⋮ Calculate the z-test statistic using the sample proportion $\hat{p}$, the null hypothesis proportion $p_0$, and the standard error SE. ‖ ‖ $z = \frac{\hat{p} - p_0}{SE}$ ‖ ‖ step 4 ⋮ Substitute the values into the z-test statistic formula and calculate. ‖ ‖ $z = \frac{\frac{48}{64} - 0.8}{\sqrt{\frac{0.8(1 - 0.8)}{64}}}$ ‖ ‖ step 5 ⋮ Simplify the calculations to find the z-test statistic. ‖ ‖ $z = \frac{0.75 - 0.8}{\sqrt{\frac{0.8 \cdot 0.2}{64}}}$ ‖ ‖ $z = \frac{-0.05}{\sqrt{\frac{0.16}{64}}}$ ‖ ‖ $z = \frac{-0.05}{0.05}$ ‖ ‖ $z = -1$ ‖ ∻Answer∻ ⚹ The test statistic for their significance test is $z = -1$. ⚹ ∻Key Concept∻ ⚹Hypothesis Testing and Z-Test Statistic⚹ ∻Explanation∻ ⚹The z-test statistic is used in hypothesis testing to determine how many standard errors a sample proportion is from the hypothesized population proportion under the null hypothesis. It is calculated by subtracting the null hypothesis proportion from the sample proportion, then dividing by the standard error of the sample proportion.⚹◊What is the formula for calculating the test statistic in a hypothesis test for the proportion of homes with high-speed internet access?⍭ Generate me a similar question◊

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