Question

A random sample of 100 people was taken. Eighty-five of the people in the sample favored Candidate A. At significance level 0.05, we are interested in performing a hypothesis test to determine whether or not the proportion of the population in favor of Candidate A is significantly more than 80%.The critical value is?

Answer 1

Answer:

We conclude that the proportion of the population in favor of Candidate A is less than or equal to 80%.

Step-by-step explanation:

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

We are interested in performing a hypothesis test to determine whether or not the proportion of the population in favor of Candidate A is significantly more than 80%.

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

SO, Null Hypothesis, $H_0$ : p $\leq$ 80%   {means that the proportion of the population in favor of Candidate A is less than or equal to 80%}

Alternate Hypothesis, $H_A$ : p > 80%   {means that the proportion of the population in favor of Candidate A is significantly more than 80%}

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, $\hat p$  = sample proportion of people favored Candidate A = $\frac{85}{100}$ = 85%

            n = sample of people = 100

So, test statistics  =  $\frac{0.85-0.80}{{\sqrt{\frac{0.85(1-0.85)}{100} } } } }$

                               =  1.40

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

Therefore, we conclude that the proportion of the population in favor of Candidate A is less than or equal to 80%.