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
[tex]H_0: p = 0.203\\H_a: p >0.203[/tex]
(Right tailed test at 5% level)
Std error = [tex]\sqrt{\frac{PQ}{n} } =0.0367[/tex]
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%
