Question

A random sample of 100 people was taken. Eighty of the people in the sample favored Candidate A. We are interested in determining whether or not the proportion of the population in favor of Candidate A is significantly more than 75%. At a .05 level of significance, it can be concluded that the proportion of the population in favor of candidate A is _____.

Answer 1

Answer:

It can be concluded that the proportion of the population in favor of candidate A is significantly less than or equal to 75%.

Step-by-step explanation:

We are given that a random sample of 100 people was taken. Eighty of the people in the sample favored Candidate A.

We are interested in determining whether or not the proportion of the population in favor of Candidate A is significantly more than 75%.

Let p = proportion of the population in favor of Candidate A

SO, Null Hypothesis, $H_0$ : p $\leq$ 75%  {means that proportion of the population in favor of Candidate A is significantly less than or equal to 75%}

Alternate Hypothesis, $H_a$ : p > 75%  {means that proportion of the population in favor of Candidate A is significantly more than 75%}

The test statistics that will be used here is One-sample z proportion statistics;

                   T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p (1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = proportion of the people who favor Candidate A in a sample of

                  100 people = $\frac{80}{100}$ = 0.80

           n  = sample of people = 100

So, test statistics = $\frac{0.80-0.75}{\sqrt{\frac{0.80 (1-0.80)}{100} } }$

                              = 1.25

Now, at 0.05 level of significance, the z table gives critical value of 1.6449. Since our test statistics is less than the critical value of z so we have insufficient evidence to reject null hypothesis as it will not fall in the rejection region.

Therefore, at a 0.05 level of significance, it can be concluded that the proportion of the population in favor of candidate A is significantly less than or equal to 75%.