Question

The worldwide market share for a web browser was 20.3​% in a recent month. Suppose that a sample of 120 random students at a certain university finds that 30 use the browser. Complete parts​ (a) through​ (d) below. a. 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 20.3​%?

Answer 1

Answer:

Step-by-step explanation:

Given that P = population proportion for the worldwide market share for a web browser was 20.3​% in a recent month.

Sample size n =120 and sample proportion p = 30/120 = 0.25=25%

p difference = 0.203-0.25 = -0.047

$H_0: p = 0.203\\H_a: p >0.203$

(Right tailed test at 5% level)

Std error = $\sqrt{\frac{PQ}{n} } =0.0367$

Test statistic Z = p diff/std error = 1.28

Since 1.28 lies in the range |z|<1.96, at 95% level we accept null hypothesis

There is  evidence that the market share for the web browser at the university is greater than the worldwide market share of 20.3​%