Question

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:

Answer 1

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:

$H_0: p \leq 0.75$

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

$H_1: p > 0.75$

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

• $\overline{p}$ 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:

$n = 100, \overline{p} = \frac{80}{100} = 0.8, p = 0.75$

Hence:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.8 - 0.75}{\sqrt{\frac{0.75(0.25)}{100}}}$

$z = 1.15$

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

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 brainly.com/question/25584945