MY Question is attached as an image below.
I have ansswered it below, but I think I've got (b) wrong, not sure how to get to the right answer.
Exercise 5.12
(a)
We are 95% sure that the true proportion of survey respondents’ answers of “not good” are between 3.4% and 4.24% with a mean of 3.82%. The mean is the midpoint of the confidence interval (3.4% + .42%).
(b)
95% confidence means that about 95% of the intervals contain the parameter, p. The Standard Error is .0011 or 1.1%.
> (sqrt(.0382*(1-.0382)/1151))*1.96
[1] 0.01107369 = Standard Error
> 1.96*.011
[1] 0.02156
> .0382+0.02156
[1] 0.05976
> .0382-0.02156
[1] 0.01664
(c)
The interval at 99% will be wider. At 99% the standard deviation from the mean will be -2.58 to +2.58 , versus -1.96 to +1.96 at the 95% confidence interval.
(d)
The standard error of estimate will be larger. A bigger sample tends to provide a more precise point estimate than a smaller sample. In this case there is a “larger net” with this smaller sample due to less than ½ as many Americans (500) participating in the survey.
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