Answer to Question #212027 in Statistics and Probability for nat

Question

N&M magazine believes that it has a 38% share of the national female readership market of women’s magazines. When 2 000 readers of women’s magazines were randomly selected and interviewed, 700 stated that they read N&M regularly.

 (3)

2.1.1Does the sample evidence support their claim? Explain.(5)

2.1.2 What is the population of interest in this case?(2)

2.1.3What is the sample in this case?

(2)

2.1.4 What percentage of readers interviewed read N&M magazine regularly? Is this a statistic or a parameter? Explain.

2.1.5 What does the value 38% represent in the above case? Explain.

Answer 1

2.1.1.

$H0:P=0.38$

$Ha:P\ne0.38$

$z=\frac{\hat P-P}{\sqrt\frac{P(1-P)}{n}}$

$=\frac{0.35-0.38}{\sqrt\frac{0.38\times0.62}{2000}}$

$=-2.764$

Let the significance level be 0.05.

Thus, the critical values are,

$z_{0.025}=-1.96, 1.96$

Since, 2.764>1.96, we reject null hypothesis. There is no sufficient evidence to support the claim that N&M magazine has a 38% share of the national female readers market of women’s magazines.

2.1.2

The population of interest is all female in the nation who are readers of women's magazines.

2.1.3

The sample is the 2000 readers of women magazines who were selected and interviewed.

2.1.4

The percentage of interviewers who read N&M magazine regularly is $\frac{700}{200}\times100=35\%$ . It is a statistic since it is calculated from sample data.

2.1.5

38% represents a parameter since it describes the entire population. It is calculated using census data. However, in this case, it is a claim.