According to a large poll in a previous yearabout 80\% of homes in a certain county had access to high-speed internet. The following yearresearchers wanted to test H 0 :p=0.8 vee cr; H_{3} / p < 0.8 where p is the proportion of homes in this county with high-speed internet access They took a random sample of 250 homes from that county and found that 189 of them had access to high speed internet. The test statistic for these results was epsilon approx-1.74 ?

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%.

What are the hypothesis tested?

The null hypothesis is:

[tex]H_0: p = 0.8[/tex]

The alternative hypothesis is:

[tex]H_1: p < 0.8[/tex]

What is the decision?

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

Alexkingcastrogo Alexkingcastrogo · Answer 2

Alexkingcastrogo Alexkingcastrogo

Answer:

P-value ≈ 0.0409

Step-by-step explanation:

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