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bearhunter [10]
3 years ago
14

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 cer

tain 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​%?
Mathematics
1 answer:
ziro4ka [17]3 years ago
4 0

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​%

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