Question

In: Statistics and Probability

In 1997, a survey of 840 households showed that 131of them use e-mail. Use those sample...

In 1997, a survey of 840 households showed that 131of them use e-mail. Use those sample results to test the claim that more than 15% of households use e-mail. Use a 0.05 significance level. Use this information to answer the following questions.

Expert Solution

Solution :

Given that,

= 0.15

1 - =0.85

n = 840

x = 131

Level of significance = = 0.05

Point estimate = sample proportion = = x / n = 131/840 = 0.156

This a right (One) tailed test.

Ho: p = 0.15

Ha: p 0.15

Test statistics

z = ( - ) / *(1-) / n

= ( 0.156 - 0.15) / (0.15*0.85) /840

= 0.49

P-value = P(Z>z)

= 1 - P(Z <z )

= 1- P(Z < 0.49)

= 1 - 0.6241

= 0.3759

The p-value is p = 0.3759, and since p = 0.3759 >0.05, it is concluded that the null hypothesis is not rejected.

Conclusion:

It is concluded that the null hypothesis Ho is not rejected. Therefore there is no enough evidence to claim that the population

proportion is greater than 0.15 at 0.05 level o0f significance.

orchestra answered 2 years ago

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