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.
a) Define Population 1 and Population 2.
b) Define the parameter of interest.
c) Name the distribution required to calculate confidence intervals. (Check the relevant criteria.)
d) Construct a 90% confidence interval for the true difference in proportions.
e) Interpret your confidence interval.
f) Is there a difference in the proportions (at a 90% CL)?
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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