Question

Yahoo did a survey to determine the percentage of people who used the Internet at work : In México it was found that 40% of adults use the Internet of a sample of 240. In Monterrey , 32% of adults. used the Internet of a sample of 250. Is the proportion that uses the Internet in Mexico greater than in Monterrey for a 95% confidence level? a) Hypothesis b) Statistic test C) P-value and graph d) Conclusion

Answer 1

Answer)

A)

Null hypothesis Ho : P1 = P2

Alternate hypothesis Ha : P1 > P2

B)

N1 = 240, P1 = 0.4

N2 = 250, P2 = 0.32

First we need to check the conditions of normality that is if n1p1 and n1*(1-p1) and n2*p2 and n2*(1-p2) all are greater than equal to 5 or not

N1*p1 = 96

N1*(1-p1) = 144

N2*p2 = 80

N2*(1-p2) = 170

All the conditions are met so we can use standard normal z table to conduct the test

Test statistics z = (P1-P2)/standard error

Standard error = √{p*(1-p)}*√{(1/n1)+(1/n2)}

P = pooled proportion = [(p1*n1)+(p2*n2)]/[n1+n2]

After substitution

Test statistics z = 1.85

From z table, P(Z>1.85) = 0.0322

C)

So, P-Value = 0.0322

D)

As the obtained P-Value is less than the given significance 0.05

We reject the null hypothesis Ho

And we have enough evidence to conclude that p1 > p2

200 rejection region -) P-velve Scanned with CamScanner