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%?

Step-by-step explanation:

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

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

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

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

The z-value is calculated as:

-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%.

could

step-by-step explanation:

u send me a picture of ur work

4

step-by-step explanation:

6 (donuts) divided by 3 is 2.

so 2/3 means 2 times 2, which is 4/6.

you get two and your friend gets four.