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olya-2409 [2.1K]
3 years ago
9

A major home improvement store conducted its biggest brand recognition campaign in the company's history. A series of new televi

sion advertisements featuring well-known entertainers and sports figures was launched. A key metric for the success of television advertisements is the proportion of viewers who "like the ads a lot." A study of 1,189 adults who viewed the ads reported that 230 indicated that they "like the ads a lot." The percentage of a typical television advertisement receiving the "like the ads a lot" score is believed to be 22%. Company officials wanted to know if there is evidence that the series of television advertisements are less successful than the typical ad (i.e. if there is evidence that the population proportion of "like the ads a lot" for the company's ads is less than 0.22) at a 0.01 level of significance. 1) State H subscript 0 and H subscript 1. (2 points) 2) What critical value should the company officials use to determine the rejection region? (2 points) 3) What is the value of the test statistic? (4 points) 4) What is the conclusion of the hypothesis test? Is there evidence that the population proportion of "like the ads a lot" for the company's ads is less than 0.22?

Mathematics
1 answer:
ollegr [7]3 years ago
3 0

Answer:

Step-by-step explanation:

Identify:      Hypothesized π = 0.22 n =1189 X = 230  

Estimate the sample proportion, p, of defective items. p= X/n = 230/1189 = 0.1934

List the formula for the standard error of a proportion.   \sqrt{PQ/n}

Estimate the standard error of a proportion.  \sqrt{PQ/n} = \sqrt{0.22\times 0.78/1189} = 0.0120

State the null and alternative hypotheses below.

H0: P = 0.22

H1: P<0.22

Illustrate the acceptance and rejection regions of the diagram to the right.

check the attached image below for the diagram

Identify the critical Z , -1.645

Fornulate the decision rule and list below.

If Z calculated is less than Z critical we reject null hypothesis

Calculate your test statistic.

Z= \frac{p-P}{\sqrt{PQ/n}} = \frac{0.1934-0.22}{0.0120}= -2.217

Do you accept or reject the null hypothesis at 0.05 level of significance? Reject

State your conclusion. Should the quality control manage accept the shipment or return it?

WE have sufficient evidence that the population proportion of "like the ads a lot" for the company's ads is less than 0.22 at a 0.05 level of significance.

If a 0.01 level of significance had been used in the test, would you have accepted or rejected the null hypothesis?  

at 0.01 level of significance the critical value of Z is -2.33

Since Z calculated is greater than Z critical value we accept the null hypothesis

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