Question

A photographer is interested in whether there is a difference in the proportions of adults and...

  1. A photographer is interested in whether there is a difference in the proportions of adults and children who like to take pictures. In a survey of 310 randomly selected adults and 270 randomly selected children, 219 adults and 215 children said they like to take pictures.

    1. a) Define Population 1 and Population 2.

    2. b) Define the parameter of interest.

    3. c) Name the distribution required to calculate confidence intervals. (Check the relevant criteria.)

    4. d) Construct a 90% confidence interval for the true difference in proportions.

    5. e) Interpret your confidence interval.

    6. f) Is there a difference in the proportions (at a 90% CL)?

Homework Answers

Answer #1

x1 = number of adults who like to take pictures = 219

n1 = total number of adults selected randomly = 310

               (Round to 4 decimal)

x2 = number of children who like to take pictures = 215

n2 = total number of children selected randomly =270

         (Round to 4 decimal)

a) Population 1: Adults

     Population 2: Children

b) Here parameter of interest is difference in the proportions of adults and children who like to take pictures.

c) Here we use normal distribution.

d) Confidence level = c = 0.90

90% confidence interval for the true difference in proportions is

Where zc is z critical value for (1+c)/2 = (1+0.90)/2 = 0.95

zc = 1.65             (From statistical table of z values)

           

             (Round to 4 decimal)

90% confidence interval for the true difference in proportions is (-0.1486, -0.0310).

e) Interpretation:

We are 90% confident that the true difference in proportions will lie between the interval (-0.1486, -0.0310).

f) Here confidence interval does not contain zero.

So we can say there is difference in the proportions at 90% confidence level.

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