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

C

Step-by-step explanation:

Solution:-

- A medical school claim was made on the population proportion that more than P = 0.28 (28%) of its student plan to go into general practise.

- A sample of n = 130 students were taken and the sample proportion was found out to be p = 0.32 ( 32% ).

- We will first estimate the sample standard deviation (σ) by assuming that the population is normally distributed with conditions :

                       n*P = 130*0.28 = 36.4 ≥ 10

                       n*( 1 - P ) = 130*0.72 = 93.6 ≥ 10

- The condition of normality are valid. The population is assumed to be normally distributed. The sample must also be normally distributed. The sample standard deviation (σ):

                        σ =  √[ P*(1-P) /n ]   = √[ 0.28*(1-0.28) /130 ]        

                        σ = √0.00155 = 0.03937          

- The Z-score test statistic for the sample proportion p can be determined by:

                       Z-test = ( p - P ) / σ

                       Z-test = ( 0.32 - 0.28 ) / 0.0397

                       Z-test = 1.00755

- The p-value of the Z-test is the probability of values where Z value is greater than the Z-test:

                   p-value = P ( Z > Z-test )

                   p-value = P ( Z > 1.00755 ) = 1 - 0.843

                                                                = 0.157

- The p -value is = 0.157 ≈ 0.1539

Answer is C. 0.1539

Refer below.

Step-by-step explanation:

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. The P-value for a test of the school's claim is:

0.1539