Mike runs for the president of the student government and is interested to know whether the proportion of the student body in fa

Question

3 months ago

7

Mike runs for the president of the student government and is interested to know whether the proportion of the student body in fa

vor of him is significantly more than 50 percent. A random sample of 100 students was taken. Fifty-five of them favored Mike. At a 0.05 level of significance, it can be concluded that the proportion of the students in favor of Mike:
a.is significantly greater than 50 percent because 55 percent of the sample favored him.
b.is not significantly greater than 50 percent.
c.is significantly greater than 55 percent.
d.is not significantly different from 55 percent.

Mathematics

1 answer:

zzz [12.3K] · Answer 1 · 2026-03-03T03:14:10+03:00

Answer:

We conclude that less than or equal to 50% of the student body supports him significantly.

Step-by-step explanation:

Mike, while campaigning for the student government presidency, is keen to determine if more than 50% of the student body supports him.

A random sample of 100 students was surveyed, with 55 expressing support for Mike.

Let p = the proportion of students backing Mike.

Thus, Null Hypothesis, $H_0$ : p $\leq$ 50% {indicating that the proportion of supporters among the student body is significantly less than or equal to 50%}

Alternate Hypothesis, $H_A$ : p > 50% {indicating that the proportion of supporters among the student body is significantly more than 50%}

The test statistics to be utilized here One-sample z proportion statistics;

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

where, $\hat p$ = the sampled proportion in favor of Mike = $\frac{55}{100}$ = 0.55

n = number of sampled students = 100

Therefore, test statistics = $\frac{0.55-0.50}{\sqrt{\frac{0.55(1-0.55)}{100} } }$

= 1.01

The z test statistic is 1.01.

At a significance level of 0.05, the z table provides a critical value of 1.645 for a right-tailed test.

Given that our test statistic falls below the critical value of z (1.01 < 1.645), we lack sufficient evidence to reject the null hypothesis as it remains outside the rejection region, leading to failure to reject our null hypothesis.

As a result, we conclude that less than or equal to 50% of the student body is in favor of him or the proportion supporting Mike is not significantly greater than 50%.