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 Ha: 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.
Problem 3 (10 marks) Based on the BBM TV ratings, the Sudbury local television claims its...
Problem 6 (10 marks) Based on the BBM TV ratings, the Sudbury local television claims its 11:00 PM newscast reaches at least 41% of the viewing audience in the area. In a survey of 100 viewers, 36 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?
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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? نیا
Based on the Nielsen ratings, the local CBS affiliate claims its 11:00 PM newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 31% indicated that they watch the late evening news on this local CBS station. The sample proportion is 0.58 0.41 0.31 0.51
Problem 5 (7 marks) A survey of 25 retail stores revealed that the average price of a DVD was $375 with a standard deviation of $20. a) What is the 95% confidence interval to estimate the true cost of the DVD? (4 marks) b) What sample size would be needed to estimate the true average price of a DVD with an error of +55 and a 99% confidence? (3 marks) Problem 6 (10 marks) Based on the BBM TV ratings,...
Problem 3 (10 marks) A candidate to the municipal elections in Sudbury claims that at least 5% of the voters favors his party. A random sample of 1000 voters in the city revealed that 40 of them were planning to vote for him. At a significance level a=0.05, can you infer that the candidate's claim is false?