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 sampled. ‖ ‖ step 2 ⋮ $\hat{p} = \frac{48}{64}$ ‖ ‖ step 3 ⋮ Simplify the sample proportion: $\hat{p} = 0.75$ ‖ ‖ step 4 ⋮ Calculate the standard error (SE) using the null hypothesis proportion ($p_0$) and the sample size ($n$): $SE = \sqrt{\frac{p_0(1-p_0)}{n}}$ ‖ ‖ step 5 ⋮ $SE = \sqrt{\frac{0.8(1-0.8)}{64}}$ ‖ ‖ step 6 ⋮ Simplify the standard error: $SE = \sqrt{\frac{0.8 imes 0.2}{64}} = \sqrt{\frac{0.16}{64}} = \sqrt{0.0025} = 0.05$ ‖ ‖ step 7 ⋮ Calculate the test statistic ($z$) 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 8 ⋮ $z = \frac{0.75 - 0.8}{0.05}$ ‖ ‖ step 9 ⋮ Simplify the test statistic: $z = \frac{-0.05}{0.05} = -1$ ‖ ∻Answer∻ ⚹ The test statistic ($z$) for this significance test is -1.00 (rounded to two decimal places). ⚹ ∻Key Concept∻ ⚹Hypothesis Testing and Test Statistic Calculation⚹ ∻Explanation∻ ⚹The test statistic in hypothesis testing measures how far the sample proportion is from the null hypothesis proportion, in standard error units. It is used to determine whether to reject the null hypothesis. In this case, the test statistic helps us understand if the proportion of homes with high-speed internet access has significantly changed from the previous year's proportion of 0.8.⚹◊What formula is used to calculate the test statistic in hypothesis testing for proportions, specifically in this scenario where the null hypothesis is $H_{0}: p = 0.8$ and the alternative hypothesis is $H_{a}: p eq 0.8$?⍭ Generate me a similar question◊

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