Give confidence intervals for the mean BMI and the margins o

Give confidence intervals for the mean BMI and the margins of error for 90%, 95%, and 99% confidence.

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Find the margins of error for 95% confidence based on SRSs of N young women.

Conf. Level	Interval (0.01)	margins of error (0.0001)
90% __________	to ________	____________
95% __________	to _________	_____________
99% _________	to ____________	_______________ We have the survey data on the body mass index ( BMI) of 645 young women. The mean BMI in the sample was . We treated these data as an SRS from a Normally distributed population with standard deviation 7.7. Find the margins of error for 95% confidence based on SRSs of N young women. N margins of error (0.0001) 143    423    1553

Solution

1.

We have the survey data on the body mass index (BMI) of 652 young women. The mean BMI in the sample was . We treated (x bar) x= 26.4 these data as an SRS from a Normally distributed population with standard deviation geteq.ashx?eqtext=%26sigma%3B%3D 7.2.
Give confidence intervals for the mean BMI and the margins of error for 90%, 95%, and 99% confidence.

a)

For 90% confidence:

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.05          
X = sample mean =    26.4          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    7.2          
n = sample size =    652          
              
Thus,              
Margin of Error E =    0.463805565 [ANSWER]
          
Lower bound =    25.93619443          
Upper bound =    26.86380557          
              
Thus, the confidence interval is              
              
(   25.93619443   ,   26.86380557   ) [ANSWER]

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b)

Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    26.4          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    7.2          
n = sample size =    652          
              
Thus,              
Margin of Error E =    0.552658418   [ANSWER]      

Lower bound =    25.84734158          
Upper bound =    26.95265842          
              
Thus, the confidence interval is              
              
(   25.84734158   ,   26.95265842   ) [ANSWER]

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c)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    26.4          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    7.2          
n = sample size =    652          
              
Thus,              
Margin of Error E =    0.726316279   [ANSWER]
      
Lower bound =    25.67368372          
Upper bound =    27.12631628          
              
Thus, the confidence interval is              
              
(   25.67368372   ,   27.12631628   ) [ANSWER]

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