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

Question

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   

Explanation / Answer

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]

*********************

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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