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Using the z-distribution, it is found that characteristics are:
a) 0.2
b) 0.0058
c) z = 2.575
d) 0.0149
e) (0.1851, 0.2149).
f) We are 99% confident the true proportion of adults who purchased something online is between 0.1851 and 0.2149.
Item a:
The point estimate is the sample proportion, hence:
[tex]\pi = \frac{789}{4750} = 0.2[/tex]
Item b:
The standard error is:
[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.2(0.8)}{4750}} = 0.0058[/tex]
Item c:
99% confidence level, hence[tex]\alpha = 0.99[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so [tex]z = 2.575[/tex].
Item d:
The margin of error is:
[tex]M = zs = 2.575(0.0058) = 0.0149[/tex]
Item e:
The interval is the point estimate plus/minus the margin of error, hence
[tex]\pi - M = 0.2 - 0.0149 = 0.1851[/tex]
[tex]\pi + M = 0.2 + 0.0149 = 0.2149[/tex]
The 99% confidence interval estimate of the population proportion of adults who had purchased something online is (0.1851, 0.2149).
Item f:
We are 99% confident the true proportion of adults who purchased something online is between 0.1851 and 0.2149.
A similar problem is given at https://brainly.com/question/16236451
