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According to a very large poll in 2015. about 90% of homes in California had access to the internet. Market researchers want to test if that proportion is now higher, so they take a random sample of 100 homes in California and find that 96 of them have access to the internet.
The researchers will test H, : P = 0.90 versus H, :p > 0.90, where p is the proportion of homes in California that have access to the internet. Assuming that the conditions for inference have been met, calculate the test statistic for their significance test.
z = ?​

Respuesta :

Using the z-distribution, as we are working with a proportion, it is found that the test statistic for their significance test is given by: z = 2.

What are the hypothesis tested?

As stated in the problem, they are:

[tex]H_0: p = 0.9[/tex]

[tex]H_1 = p > 0.9[/tex]

What is the test statistic?

It is given by:

[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

In which:

  • [tex]\overline{p}[/tex] is the sample proportion.
  • p is the proportion tested at the null hypothesis.
  • n is the sample size.

In this problem, we have that;

[tex]n = 100, \overline{p} = \frac{96}{100} = 0.96, p = 0.9[/tex]

Hence:

[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

[tex]z = \frac{0.96 - 0.9}{\sqrt{\frac{0.9(0.1)}{100}}}[/tex]

[tex]z = 2[/tex]

The test statistic for their significance test is given by: z = 2.

More can be learned about the z-distribution at https://brainly.com/question/16313918

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