Help please 1 In a survey of 634 males ages 1864 396 say the

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

Solution

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]

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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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