A medical school claims that more than 28% of its students plan to go into general practice. It is found that among a random sample of 130 of the school's students, 32% of them plan to go into general practice. Find the P-value for a test of the school's claim.

a. 0.1635
b. 0.3461
c. 0.1539
d. 0.3078

And the best option would be:

c. 0.1539

Step-by-step explanation:

Data given and notation

n=130 represent the random sample taken

estimated proportion of students who plan to go into general practice

is the value that we want to test

z would represent the statistic (variable of interest)

represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.28:  

Null hypothesis:  

Alternative hypothesis:  

When we conduct a proportion test so we need to use the z statistic, and the is given by:  

(1)  

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

Statistical decision  

The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

And the best option would be:

c. 0.1539

she played for 18.75 hours.

step-by-step explanation:

if she played 2/4 of a hour on tuesday. and there is 60 minutes in an hour and 60 divided by 4 is 7.5. and she played 2/4 of an hour on thursday then again 60 divided by 4 is 7.5, you add 7.5+7.5=56.25. now you have to do 1/4 of an hour, it equals 3.75. now you have to add them all up. 3.75+7.5+7.5=18.75

i hope this and i hope you get it right. it took me a while to get this done for you. : )