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Refer to previous question. At a .05 level of significance, it can be concluded that the proportion of the population in favor of candidate A is:

Sagot :

Using the z-distribution, it is found that since the test statistic is less than the critical value for the right-tailed test, it is found that it can be concluded that the proportion of the population in favor of candidate A is not significantly greater than 0.75.

At the null hypothesis, it is tested if the proportion is not significantly more than 75%, that is:

[tex]H_0: p \leq 0.75[/tex]

At the alternative hypothesis, it is tested if the proportion is significantly more than 75%, that is:

[tex]H_1: p > 0.75[/tex]

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.

In the sample, 80 out of 100 people favored candidate A, hence, the parameters are:

[tex]n = 100, \overline{p} = \frac{80}{100} = 0.8, p = 0.75[/tex]

Hence:

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

[tex]z = \frac{0.8 - 0.75}{\sqrt{\frac{0.75(0.25)}{100}}}[/tex]

[tex]z = 1.15[/tex]

The critical value for a right-tailed test, as we are testing if the proportion is greater than a value, is [tex]z^{\ast} = 1.645[/tex].

Since the test statistic is less than the critical value for the right-tailed test, it is found that it can be concluded that the proportion of the population in favor of candidate A is not significantly greater than 0.75.

To learn more about the use of the z-distribution to test an hypothesis, you can take a look at https://brainly.com/question/25584945