Question

Problem 2
A consumer products firm has recently introduced a new brand. The firm would like to estimate the proportion of people in its target market segment who are aware of the new brand. As part of a larger market research study, it was found that in a sample of 125 randomly selected individuals from the target market segment, 84 individuals were aware of the firm’s new brand.

a. Construct a 95% confidence interval for the proportion of individuals in the target market segment who are aware of the firm’s new brand. If you use Excel and/or StatTools, please specify any functions you use and all the inputs.

b. The manager in charge of the new brand has stated that the brand awareness is greater than .75, meaning that more than 75% of the population is aware of the brand. He would like to use hypothesis testing to prove his claim. At the 5% significance level, conduct a hypothesis test with the goal of proving his claim. In particular
i) specify the null hypothesis and the alternative hypothesis,
ii) state whether you are using a one- or two-tailed test,
iii) specify the p-value of the test, and iv) provide the results of the test in “plain English”.
Hints: 1) Think carefully about what is the null hypothesis, and what is the alternative. 2) If you want to use StatTools for the analysis, you need to create the survey data in Excel first; then you can run the analysis.

For future polls, the firm is interested in minimizing their marketing-research costs. The margin of error (MOE) in their pools should be no larger than B (i.e., ± 100B percentage points). To simplify the analysis, we will assume their marketing study only has a single question that is used to estimate p, at the 5% significance level. I

c. If the firm had no prior knowledge of p, how large would the sample size have to be to ensure MOE ≤ B? (Hint: your answer will be a formula containing B).
d. The firm has prior knowledge that p will likely be somewhere between .10 and .20 (that is between 10% and 20%). How large should the sample be if they would like to ensure that MOE (=B) is no larger than 0.01 or 1%?


*Please show work in Excel. I have almost entirely solved a and b, and mostly need help with c & d. Thanks

Answer 1

Question 1

Arrangement:

We are given

Test measure = n = 125

Number of achievements = x = 84

Certainty level = 95%

Test extent = P = x/n = 84/125 = 0.672

Basic z esteem = 1.96 (by utilizing z-table or exceed expectations)

Certainty interim = P -/+ Z*sqrt(P*(1 – P)/N)

Certainty interim = 0.672 -/+ 1.96*sqrt(0.672*(1 – 0.672)/125)

Certainty interim = 0.672 -/+ 1.96*0.0420

Certainty interim = 0.672 -/+ 0.0823

Lower limit = 0.672 – 0.0823 = 0.5897

Furthest breaking point = 0.672 + 0.0823 = 0.7543

Certainty interim = (0.5897, 0.7543)

Question 2

Here, we need to utilize z test for populace extent. The invalid and elective theory for this test is given as beneath:

(I)

Invalid speculation: H0: The extent of the people who know about new brand is 75%.

Elective theory: Ha: The extent of the people who know about new brand is more noteworthy than 75%.

(ii)

H0: p = 0.75 versus Ha: p > 0.75

This is a one followed test. (Upper followed or right followed test)

We are given

Test measure = n = 125

Number of achievements = x = 84

Test extent = P = x/n = 84/125 = 0.672

Dimension of criticalness = α = 5% = 0.05

Basic Z esteem = 1.6449

Test measurement equation is given as beneath:

Z = (P – p)/sqrt(p*(1 – p)/n)

Z = (0.672 – 0.75)/sqrt(0.75*(1 – 0.75)/125)

Z = - 0.078/0.0387

Z = - 2.0140

(iii)

P-esteem = 0.9780

(by utilizing z-table or exceed expectations)

(Exceed expectations direction: =1-normsdist(- 2.0140))

α = 0.05

P-esteem > α

In this way, we don't dismiss the invalid speculation that the extent of the people who know about new brand is 75%.

(iv)

There is inadequate proof to reason that extent of the people who know about new brand is more prominent than 75%.