Question

A research report claims that 20% of all individuals use Firefox to browse the web. A software company is trying to determine if the proportion of their users who use Firefox is significantly different from 0.2. In a sample of 200 of their users, 32 users stated that they used Firefox. Using this data, conduct the appropriate hypothesis test using a 0.05 level of significance.
a) What are the appropriate hypotheses?
H0: p = 0.2 versus Ha: p < 0.2
H0: p = 0.2 versus Ha: p > 0.2
H0: p = 0.2 versus Ha: p ? 0.2
H0: ? = 0.2 versus Ha: ? > 0.2
b) What is the test statistic? Give your answer to four decimal places.
c) What is the P-value for the test? Give your answer to four decimal places.
d) What is the appropriate conclusion?
1. Conclude that the Firefox proportion is not 0.2 because the P-value is smaller than 0.05.
2. Conclude that the Firefox proportion is not 0.2 because the P-value is larger than 0.05.
3. Fail to reject the claim that the Firefox proportion is 0.2 because the P-value is larger than 0.05.
4. Fail to reject the claim that the Firefox proportion is 0.2 because the P-value is smaller than 0.05.

Answer 1

Answer:

a) The null and alternative hypothesis are:

$H_0: \pi=0.2\\\\H_a:\pi\neq 0.2$

b) Test statistic z = -1.3258

c) P-value = 0.1849

Step-by-step explanation:

a) As the reseracher want to determine  if the proportion of their users who use Firefox is significantly different from 0.2, it is a two-tailed test, as a proportion significantly low or significantly high gives evidence to support the alternative hypothesis.

Then the null and alternative hypothesis are:

$H_0: \pi=0.2\\\\H_a:\pi\neq 0.2$

b)  The sample has a size n=200.

The sample proportion is p=0.16.

$p=X/n=32/200=0.16$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.2*0.8}{200}}\\\\\\ \sigma_p=\sqrt{0.0008}=0.028$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.16-0.2+0.5/200}{0.028}=\dfrac{-0.038}{0.028}=-1.3258$

c) This test is a two-tailed test, so the P-value for this test is calculated as:

$\text{P-value}=2\cdot P(z<-1.3258)=0.1849$

As the P-value (0.1849) is greater than the significance level (0.05), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the proportion of their users who use Firefox is significantly different from 0.2