Question

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, 39% of them plan to go into general practice. Find the P-Value for a test of the school's claim.

0.9974
0.1635
0.3078
0.0026

Answer 1

Answer: 0.0026

Step-by-step explanation:

Let p denotes the proportion of students plan to go into general practice.

As per given , we have

Alternative hypothesis : $H_a: p>0.28$

Since the alternative hypothesis $(H_a)$ is right-tailed so the test is  a right-tailed test.

Also , it is given that ,

i.e. sample size : = 130

x= 490

$\hat{p}=0.39$

Test statistic(z) for population proportion :

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

, where p=population proportion.

$\hat{p}$= sample proportion

n= sample size.

$z=\dfrac{0.39-0.28}{\sqrt{\dfrac{0.28(1-0.28)}{130}}}\\\\=\dfrac{0.11}{0.0393798073988}=2.79330975101\approx2.79$

P-value for right-tailed test = P(z>2.79)=1-P(z≤ 2.79)  [∵P(Z>z)=1-P(Z≤z)]

=1- 0.9974=0.0026  [using z-value table]

Hence, the  P-Value for a test of the school's claim = 0.0026