Question

8. In a recent poll on education 64% of all college students stay up all night studying for final exams. A professor at a local college says this is too low and wants to test this claim. He surveys 5,000 college students and finds that 3250 have stayed up all night studying for final exams. Which statement can we conclude. Use a -5% a State the null and alternative hypothesis b. Determine and draw the hypothesis testing model label everything) c. Determine the critical value(s) and test statistic d. Determine the conclusion

Answer 1

The number of students surveyed n=5000

The number of students who stayed up all night x=3250.

The proportion of students who stayed up all night $\hat{p}=\frac{x}{n}=\frac{3250}{5000}=0.65$

a. The Null and alternative hypothesis:

Ho: p=0.64

$H_{1}:p>0.64$

Level of significance:

g= 0,05

Test statistic: $Z=\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}$ where $p_{0}$ is the hypothesized proportion.

  $Z=\frac{0.65-0.64}{\sqrt{\frac{0.64(1-0.64)}{5000}}}$

$Z=\frac{0.01}{\sqrt{\frac{0.64(0.36)}{5000}}}$

  $Z=\frac{0.01}{\sqrt{\frac{0.2304}{5000}}}$

$Z=\frac{0.01}{\sqrt{0.00004608}}$

$Z=\frac{0.01}{0.006788}$

$Z=1.4731$

(c). The critical value of the test: This is an one sided test . We can look at the right tail value of a Normal distribution at 5% level of significance which is 1.6449.

(d). Since the calculated value of Z<the critical value of Z, we fail to reject the null hypothesis. Hence, we conclude that there is no sufficient evidence to claim that the professor's assertion of 64% is too low.