Question

Problem 3 (10 marks) Based on the BBM TV ratings, the Sudbury local television claims its 11:00 PM newscast reaches exactly 41% of the viewing audience in the area. In a survey of 100 viewers, 45 indicated that they watch the late evening news on this Sudbury local television. At a significance level a=0.05, can you infer that the Sudbury local television claim is false? دیا

Answer 1

Solution:

Here, we have to use one sample z test for the population proportion.

The null and alternative hypotheses for this test are given as below:

H₀: p = 0.41 versus H_a: p ≠ 0.41

This is a lower tailed test.

We are given

Level of significance = α = 0.05

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

x = number of items of interest = 45

n = sample size = 100

p̂ = x/n = 45/100 = 0.45

p = 0.41

q = 1 - p = 0.59

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.45 - 0.41)/sqrt(0.41*0.59/100)

Z = 0.8133

Test statistic = 0.8133

P-value = 0.4161

(by using z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that 11:00 PM newscast reaches 41% of the viewing audience in the area.

At a significance level α = 0.05, we cannot infer that the Sudbury local television claim is false.