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:

H_{0}: 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**

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