Question

A publisher reports that 42% of their readers own a laptop. 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 35% of the readers owned a laptop. Find the value of the test statistic. Round your answer to two decimal places.

Answer 1

Null hypothesis: H0 = 0.42

Alternative hypothesis: Ha <> 0.42

z score = −1.849 = -1.85

P value = P(Z<-1.849) + P(Z>1.849) = 0.032 + 0.032= 0.064

Explanation:

Given;

n=170 represent the random sample taken

Null hypothesis: H0 = 0.42

Alternative hypothesis: Ha <> 0.42

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 = 0.42

p^ = Observed proportion = 0.35

Substituting the values we have

z = (0.35-0.42)/√{0.42(1-0.42)/170}

z = -1.849

z = -1.85

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

P value = P(Z<-1.849) + P(Z>1.849) = 0.032 + 0.032= 0.064

Since z at 0.05 significance level is between -1.96 and +1.96, and the z score for the test (z = -1.849) 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 decimal place is 1.91

Explanation:

P cap= 0.42

n= 170

P= 0.35

q= 1-p

q= 1-0.35=0.65

Z=( pcap - p)/√(p*q)/n

Z= (0.42 - 0.35)/ √ (0.35*0.65)/170

Z= 0.07/√1.3383 * 10 ^ -3

Z= 1.91

Therefore the test statistic is z=1.91

See attached picture