Question

.An e-commerce research company claims at most59%of graduate students have bought merchandise on-line. A consumer group is suspicious of the claim and wants to test it. They collected a random sample of 135graduate student and found that50 graduate students have not bought merchandise on-line. Is there enough evidence to show that the true proportion is at most 59%? Conduct the test at 0.1level of significance. (Please show all 5 steps involved

Answer 1

Solution:

Claim: p is at most 59%

i.e. p 0.59

So , the null and alternative hypothesis are

H₀ : p 0.59 vs H_a: p > 0.59

Now,

sample size n = 135

Number of student have not bought merchandise on-line is 50

So ,

Number of student have bought merchandise on-line is 135 - 50 = 85

Let   be the sample proportion.

= x/n = 85/135 = 0.6296

The test statistic z is

z =   

=  (0.6296 - 0.59)/[0.59*(1 - 0.59)/135]

= 0.94

Test statistic z = 0.94

Right tailed test

so .

p value = P(Z > z) = 1 - P(Z < z) = 1 - P(Z < 0.94) = 1 - 0.8264 = 0.1736

p value = 0.1736

p value is greater than level of significance 0.1

So , Do not reject the null hypothesis.

There is not sufficient evidence to reject the claim.