Question

A 2010 survey by a reputable automotive website found that

6868?%

of vehicle owners avoided automotive maintenance and repairs. Suppose a company would like to perform a hypothesis test to challenge this finding. From a random sample of

110110

vehicle? owners, it was found that

8484

avoid maintenance repairs. Using

alpha?equals=0.020.02?,

answer parts a and b below.

a. What conclusions can be drawn concerning the proportion of vehicle owners who avoid automotive maintenance and? repairs?

Determine the null and alternative hypotheses. Choose the correct answer below.

A.

Upper H 0H0?:

pless than or equals?0.680.68

Upper H 1H1?:

pgreater than>0.680.68

B.

Upper H 0H0?:

pgreater than>0.680.68

Upper H 1H1?:

pless than or equals?0.680.68

C.

Upper H 0H0?:

pgreater than or equals?0.680.68

Upper H 1H1?:

pless than<0.680.68

D.

Upper H 0H0?:

pequals=0.680.68

Upper H 1H1?:

pnot equals?0.680.68

Determine the critical? value(s) of the test statistic.

z Subscript alphaz?equals=nothing

?(Use a comma to separate answers as needed. Round to

twotwo

decimal places as? needed.)

Calculate the test statistic.

z Subscript pzpequals=nothing

?(Round to two decimal places as? needed.)

Determine the conclusion. Choose the correct answer below.

A.

RejectReject

Upper H 0H0.

There

isis

sufficient evidence to conclude that the true proportion of owners who avoid maintenance and repairs is different from what the website found.

B.

Do not rejectDo not reject

Upper H 0H0.

There

is notis not

sufficient evidence to conclude that the true proportion of owners who avoid maintenance and repairs is different from what the website found.

C.

Do not rejectDo not reject

Upper H 0H0.

There

isis

sufficient evidence to conclude that the true proportion of owners who avoid maintenance and repairs is different from what the website found.

D.

RejectReject

Upper H 0H0.

There

is notis not

sufficient evidence to conclude that the true proportion of owners who avoid maintenance and repairs is different from what the website found.

b. Determine the? p-value for this test.

?p-valueequals=nothing

?(Round to three decimal places as? needed.)

Answer 1

As we are trying to test whether the proportion is 0.68, therefore the null and the alternate hypothesis here would be given as:

For 0.02 level of significance, we get from the standard normal tables:

P( -2.326 < Z < 2.326 ) = 0.98

Therefore the critical values here are: 2.326?

The sample proportion here is computed as:

p = 84 / 110 = 0.76

Now the test statistic here is computed as:

Now as the critical value here is 2.326 > 1.8805, therefore the test is not significant and we cannot reject the null hypothesis here.

Do not reject H_o is the required answer here.

The p-value for this two tailed test here is computed from the standard normal tables as:

p = 2P( Z > 1.8805 ) = 0.03*2 = 0.060

Therefore 0.060 is the p-value required here.