Question

At a certain university, 22% of the students fail general chemistry on their first attempt. Professor Brown teacher at this university and believes that the rate of first-time failure in his general chemistry classes is 45%. He samples 86 students from last semester who were first-time enrollees in general chemistry and finds that 19 of them failed his course.
1) State the appropriate null and alternate hypotheses.
2) Compute the P-value.
3) Using a = 0.05, can Professor Brown conclude that the percentage of failures differs from 45%?

Answer 1

Answer:

$a. \ H_o:p=0.45, \ \ \ \ H_a:p\neq 0.45\\\\b.\ \hat p=0.2209\\\\c. \ Yes\ (-4.2706$

Step-by-step explanation:

a. The professor's claim is that 45% first-timers fail his test. The null hypothesis is therefore stated as:

$H_o:p=0.45$

-The alternative hypothesis is that more or less people fail the test as opposed to the professor's exact claim, hence:

$H_a:p\neq 0.45$

b. To compute the P-value we use the z-value for a 95% confidence level:

$z_{\alpha/2}=z_{0.025}=1.96$

#The proportion of failures in the sample of 86 is 19:

$\hat p=\frac{19}{86}\\\\=0.2209$

The z-value is calculated as:

$z=\frac{\hat p-p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}}$

$=\frac{0.2209-0.45}{\sqrt{\frac{0.45(1-0.45)}{86}}}\\\\\\=-4.2706$

-4.2706 is less than the stated confidence level for the given 45% proportion and greatly differs from it.

- Reject the null hypothesis as there is enough evidence to reject the claim.

-Hence,Yes, Professor Brown can conclude that percentage  of failures differs from 45%.