Question

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

Answer 1

Answer:

$p_v =P(z>1.02)=0.1539$  

And the best option would be:

c. 0.1539

Step-by-step explanation:

Data given and notation

n=130 represent the random sample taken

$\hat p=0.32$ estimated proportion of students who plan to go into general practice

$p_o=0.28$ is the value that we want to test

z would represent the statistic (variable of interest)

$p_v$ 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:$p\leq 0.28$  

Alternative hypothesis:$p > 0.28$  

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

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

Calculate the statistic  

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

$z=\frac{0.32 -0.28}{\sqrt{\frac{0.28(1-0.28)}{130}}}=1.02$  

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:  

$p_v =P(z>1.02)=0.1539$  

And the best option would be:

c. 0.1539

Answer 2

Using the z-distribution, it is found that the p-value is given by:

c. 0.1539

At the null hypothesis, it is tested if the proportion is of 28% or less, that is:

$H_0: p \leq 0.28$

At the alternative hypothesis, it is tested if it is more than 28%, that is:

$H_1: p > 0.28$

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

• $\overline{p}$ is the sample proportion.

• p is the proportion tested at the null hypothesis.

• n is the sample size.

In this problem, the parameters are:

$p = 0.28, \overline{p} = 0.32, n = 130$.

Hence, the value of the test statistic is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.32 - 0.28}{\sqrt{\frac{0.28(0.72)}{130}}}$

$z = 1.01575$

The p-value is found using a z-distribution calculator, with a right-tailed test, as we are testing if the mean is more than a value, with z = 1.01575, hence it is of 0.1539, hence option c is correct.

A similar problem is given at https://brainly.com/question/17062923