Answer:
We conclude that seniors skip more than 2% of their classes at 0.01 level of significance.
Step-by-step explanation:
We are given that a professor wishes to discover if seniors skip more classes than freshmen. Suppose he knows that freshmen skip 2% of their classes.
He randomly samples a group of seniors and out of 2521 classes, the group skipped 77.
<u><em /></u>
<u><em>Let p = percentage of seniors who skip their classes.</em></u>
So, Null Hypothesis,
: p
2% {means that seniors skip less than or equal to 2% of their classes}
Alternate Hypothesis,
: p > 2% {means that seniors skip more than 2% of their classes}
The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;
T.S. =
~ N(0,1)
where,
= sample proportion of seniors who skipped their classes =
n = sample of classes = 2521
So, <u><em>test statistics</em></u> = ![\frac{\frac{77}{2521} -0.02}{{\sqrt{\frac{\frac{77}{2521}(1-\frac{77}{2521})}{2521} } } } }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7B77%7D%7B2521%7D%20-0.02%7D%7B%7B%5Csqrt%7B%5Cfrac%7B%5Cfrac%7B77%7D%7B2521%7D%281-%5Cfrac%7B77%7D%7B2521%7D%29%7D%7B2521%7D%20%7D%20%7D%20%7D%20%7D)
= 3.08
The value of the test statistics is 3.08.
Now at 0.01 significance level, <u>the z table gives critical value of 2.3263 for right-tailed test</u>. Since our test statistics is more than the critical value of z as 2.3263 < 3.08, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.
Therefore, we conclude that seniors skip more than 2% of their classes.