Question

Canadian Dial Up, an online service provider, wants to know if more than 70% of their customers spend at least 10 hours per week working online. They conducted a survey of 250 of their customers and found that 141 of them spend at least 10 hours per week working online. Test the claim using alpha=0.10.

p-value=2.70, evidence not support claim

p-value=1.000, evidence not support claim

p-value=1.350, evidence support claim

p-value=0.000, evidence support claim

p-value=0.564, evidence support claim

Answer 1

Solution :

Given that,

= 0.70

1 - = 0.30

n = 250

x = 141

Level of significance = = 0.10

Point estimate = sample proportion = = x / n = 0.564

This a right (One) tailed test.

The null and alternative hypothesis is,

Ho: p = 0.70

Ha: p 0.70

Test statistics

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

= ( 0.564 - 0.70) / (0.70*0.30) / 250

= -4.692

P-value = P(Z > z )

= 1 - P(Z < -1.692 )

= 1 - 0

= 1.000

The p-value is p = 1, and since p = 1 > 0.10, it is concluded that fail to reject the null hypothesis .

The correct option is,

p-value=1.000, evidence not support claim