Help please, 1. In a survey of 634 males ages 18-64, 396 say they have gone to t
ID: 3437679 • Letter: H
Question
Help please,
1. In a survey of 634 males ages 18-64, 396 say they have gone to the dentist in the past year. Construct 90% and 95% confidence intervals for the population. Interpret the results and compare the widths of the confidence intervals. Round to three decimal places as needed.
2. In a survey of 3385 adults, 1439 say they have started paying bills online in teh last year. Construct a 99% confidence interfal for the population interpret the results.
3. In a survey of 9000 women, 6431 say they change their nail polish once a week. Construct a 95% confidence interval for the population portion of women who change their nail polish once a week. Round three decimal places as needed.
TIA
Explanation / Answer
1.
Note that
p^ = point estimate of the population proportion = x / n = 0.624605678
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.019231014
Now, for the critical z,
alpha/2 = 0.05
Thus, z(alpha/2) = 1.644853627
Thus,
lower bound = p^ - z(alpha/2) * sp = 0.592973475
upper bound = p^ + z(alpha/2) * sp = 0.656237881
Thus, the confidence interval is
( 0.592973475 , 0.656237881 ) [ANSWER, 90% CONFIDENCE]
*********************
FOR 95% CONFIDENCE:
Now, for the critical z,
alpha/2 = 0.025
Thus, z(alpha/2) = 1.959963985
Thus,
lower bound = p^ - z(alpha/2) * sp = 0.586913583
upper bound = p^ + z(alpha/2) * sp = 0.662297773
Thus, the confidence interval is
( 0.586913583 , 0.662297773 ) [ANSWER, 95% CONFIDENCE]
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