Respuesta :
Answer:
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:
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
