Question

A publisher reports that 72% of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 170 found that 66% of the readers owned a particular make of car. Find the value of the test statistic. Round your answer to two decimal places.

Answer 1

Null hypothesis: H0 = 0.72

Alternative hypothesis: Ha <> 0.72

z score = −1.74

P value = P(Z<-1.74) + P(Z>1.74) = 0.041 + 0.041= 0.082

Explanation:

Given;

n=170 represent the random sample taken

Null hypothesis: H0 = 0.72

Alternative hypothesis: Ha <> 0.72

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 170

po = Null hypothesized value = 0.72

p^ = Observed proportion = 0.66

Substituting the values we have

z = (0.66-0.72)/√{0.72(1-0.72)/170}

z = −1.74233013109

z = -1.74

To determine the p value (test statistic) at 0.05 significance level, using a two tailed hypothesis. Using the p- value table;

P value = P(Z<-1.74) + P(Z>1.74) = 0.041 + 0.041= 0.082

Since z at 0.05 significance level is between -1.96 and +1.96, and the z score for the test (z = -1.74) falls with the region bounded by Z at 0.05 significance level. Then we can conclude that we don't have enough evidence to FAIL or reject the null hypothesis, and we can say that at 5% significance level the null hypothesis is valid.

Answer 2

The value of the test statistic to 2 d.p is z= 1.65

Explanation:

P cap= 0.72

n= 170

P= 0.66

q= 1- p

q= 1- 0.66

q= 0.34

Z=( p cap - p)/√(p*q)/n

Z= (0.72- 0.66)/√(0.66*0.34)/170

Z= 0.06/0.036332

Z= 1.65

See attached picture