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A Pew Research study finds that 23% of Americans use only a cell phone, and no land line, for making phone calls (The Wall Street Journal, October 14, 2010). A year later, a researcher samples 200 Americans and finds that 51 of them use only cell phones for making phone calls. Test whether the proportion of Americans who solely use cell phones to make phone calls differs from 23%. Use the .05 level of significance.

Respuesta :

Using the z-distribution, it is found that since the p-value is greater than 0.05, the proportion does not differ from 23%.

What are the hypotheses tested?

At the null hypotheses, it is tested if the proportion is of 23%, that is:

[tex]H_0: p = 0.23[/tex]

At the alternative hypotheses, it is tested if the proportion differs from 23%, hence:

[tex]H_1: p \neq 0.23[/tex]

What is the test statistic?

The test statistic 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.

The parameters are given as follows:

[tex]p = 0.23, n = 200, \overline{p} = \frac{51}{200} = 0.255[/tex]

The value of the test statistic is:

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

[tex]z = \frac{0.255 - 0.23}{\sqrt{\frac{0.23(0.77)}{200}}}[/tex]

z = 0.84.

Using a calculator and considering a two-tailed test, as we are testing if the proportion is different of a value, with z = 0.84, the p-value is of 0.4.

Since the p-value is greater than 0.05, the proportion does not differ from 23%.

More can be learned about the z-distribution at https://brainly.com/question/25890103

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