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

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

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{(\hat p(1-\hat p))/(n) } } } }` ~ N(0,1)

where,
`\hat p` = sample proportion of people favored Candidate A =
`(85)/(100)` = 85%

n = sample of people = 100

So, test statistics =
`\frac{0.85-0.80}{{\sqrt{(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%.