Question

A manufacturer of TV sets claims that at least 98% of its TV sets can last more than 10 years without needing a single repair. In order to verify and challenge this claim, a consumer group randomly selected 800 consumers who had owned a TV set made by this manufacturer for 10 years. Of these 800 consumers, 60 said that their TV sets needed some repair at least once.

a. Is there significant evidence showing that the manufacturer’s claim is false? Test using α = 0.01.

b. Do the data support that the manufacturer’s actual no-repair rate does not even reach 94%? Use α = 0.01.

Answer 1

Answer:

a)

Here 60 clients need at any rate one fix so 800 - 60 = 740 didn't require not fix

Given data,

n = 800,

p^ = (800 - 60) / 800

= 740/800

p^ = 0.925

Now hypothesis can be given as follows

Null hypothesis Ho : p >= 0.98

Alternative hypothesis can be given as follows

Alternative hypothesis Ha : p < 0.98

Standard deviation = sqrt(pq/n)

= sqrt(0.98*(1-0.98)/800)

= sqrt(0.0196/800)

= sqrt(2.45*10^-5)

Standard deviation = 0.0049

Test statistics will be:

z = (p^ - p) /

= (0.925 - 0.98) / 0.0049

= -0.055/0.0049

z = - 11.224

Alternative hypothesis demonstrates that the test is left followed so p-estimation of the test is

P(z < - 11.224) = 0.0000 [since from z table]

Since p-estimation of the test is under(<) 0.01 so we neglect to Ho at 0.01 significance level.

So dependent on this example, there is critical proof demonstrating that the maker's case is false.

(b)

Here 60 clients need at any rate one fix so 800-60=740 didn't require not fix.

Here we have following data:

n = 800,

p^ = (800-60)/800

p^ = 0.925

Theories are:

Ho : p >= 0.94

Ha : p < 0.94

Standard deviation of the extent is:

= sqrt(pq/n)

= sqrt(0.94(1-0.94)/800)

= sqrt(7.05*10^-5)

= 0.0084

Test statistics will be:

z = (p^ - p) /

= (0.925 - 0.94)/0.0084

= - 0.015/0.0084

z = -1.786

Alternative hypothesis Ha demonstrates that the test is left followed so p-estimation of the test is

P(z < - 1.786) = 0.0370496 [since from z table]

p value = 0.0371

Since p-estimation of the test is greater than 0.01 so we neglect the null hypothesis at 0.01 significance level.

So the manufacturer’s actual no-repair rate reach 94%