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%?
Accepted Solution
A:
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 = 1.28Since 1.28 lies in the range |z|<1.96, at 95% level we accept null hypothesisThere is evidence that the market share for the web browser at the university is greater than the worldwide market share of 20.3%