We are given that Mike runs for the president of the student government and is interested to know whether the proportion of the student body in favor of him is significantly more than 50 percent.

A random sample of 100 students was taken. Fifty-five of them favored Mike.

Let p = proportion of the students who are in favor of Mike.

So, Null Hypothesis, $H_0$ : p $\leq$ 50% {means that the proportion of student body in favor of him is significantly less than or equal to 50%}

Alternate Hypothesis, $H_A$ : p > 50% {means that the proportion of student body in favor of him is significantly more than 50%}

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

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

where, $\hat p$ = sample proportion of students body in favor of Mike = $\frac{55}{100}$ = 0.55

n = sample of students taken = 100

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

= 1.01

The value of z test statistics is 1.01.

Now, at 0.05 significance level the z table gives critical value of 1.645 for right-tailed test.

Since our test statistic is less than the critical value of z as 1.01 < 1.645, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the proportion of student body in favor of him is significantly less than or equal to 50% or proportion of the students in favor of Mike is not significantly greater than 50 percent.

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3 years ago