Using the z-distribution, it is found that since the test statistic is less than the critical value for the hypothesis test, there is enough evidence to reject the null hypothesis, that is, to conclude that the proportion of homes in the county that had access to high-speed internet is less than 80%.
The null hypothesis is:
[tex]H_0: p = 0.8[/tex]
The alternative hypothesis is:
[tex]H_1: p < 0.8[/tex]
Considering that we are working with a proportion, the z-distribution is used. Considering a standard significance level of 0.05 and a left-tailed test, as we are testing if the proportion is less than a value, the critical value is [tex]z^{\ast} = -1.645[/tex].
The test statistic is [tex]z \approx -1.74[/tex]. Since the test statistic is less than the critical value for the hypothesis test, there is enough evidence to reject the null hypothesis, that is, to conclude that the proportion of homes in the county that had access to high-speed internet is less than 80%.
To learn more about the z-distribution, you can take a look at https://brainly.com/question/26461448
Answer:
P-value ≈ 0.0409
Step-by-step explanation:
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