Question

Women athletes at the a certain university have a long-term graduation rate of 67%. Over the past several years, a random sample of 42 women athletes at the school showed that 23 eventually graduated. Does this indicate that the population proportion of women athletes who graduate from the university is now less than 67%? Use a 1% level of significance.

Answer 1

Answer: No

Explanation:

By considering the given information , the null and the alternative hypotheses will be :-

`H_0: p=0.67\\\\H_a: p<0.67`

Since the alternative hypotheses is left-tailed , so the test applied here is left-tailed.

Sample size : n= 42

The sample proportion of graduated :
`\hat{p}=(23)/(42)\approx0.55`

The test statistic for population proportion :-

`z=\frac{\hat{p}-p}{\sqrt{(p(1-p))/(n)}}`

i.e.
`z=\frac{0.55-0.67}{\sqrt{(0.67(1-0.67))/(42)}}\approx-1.65`

The p-value :
`P(z<-1.645)=0.0494715`

Since the p-value is greater than the significance level, then we do not reject the null hypothesis.

Hence, it does not give enough evidence to indicate that the population proportion of women athletes who graduate from the university is now less than 67%.

Answer 2

The null and alternative hypotheses are:

`H_(0): p=0.67`

`H_(a) : p <0.67`

Under the null hypothesis, the test statistic is:

`z=\frac{\hat{p}-p}{\sqrt{(p(1-p))/(n) } }`

Where:

`\hat{p}= (23)/(42) = 0.5476`

`\therefore z = \frac{ 0.5476 -0.67}{\sqrt{(0.67(1-0.67))/(42) }}`

`=(-0.1224)/(0.072555398)`

`=-1.69`

Now, we have to find the left tailed z critical value at 0.01 significance level using the standard normal table. The z critical value at 0.01 significance level is:

`z_(critical) = -2.33`

Conclusion: Since the test statistic does not fall in the critical region, therefore, we fail to reject the null hypothesis, and conclude that there is not sufficient evidence to say that the population proportion of women athletes who graduate from the university is now less than 67%.