Question

The worldwide market share for a web browser was 20.5​% in a recent month. Suppose that a sample of 200200 random students at a certain university finds that 5050 use the browser.

Suppose that a sample of n=800 students at the same university​ (instead of n=200​) determines that 25​% of the sample use the web browser. At the 0.05 level of​ significance, is there evidence that the market share for the web browser at the university is greater than the worldwide market share of 20.5​%? Calculate the test statistic for the second sample.

Answer 1

Solution:

Here, we have to use one sample z test for the population proportion.

The null and alternative hypotheses for this test are given as below:

H₀: p = 0.205 versus H_a: p > 0.205

This is an upper tailed test.

We are given

Level of significance = α = 0.05

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

n = sample size = 800

p̂ = x/n = 0.25

p = 0.205

q = 1 - p = 0.795

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.25 - 0.205)/sqrt(0.205*0.795/800)

Z = 3.1528

Test statistic = 3.1528