Question

Assume a 2015 Gallup Poll asked a national random sample of 491 adult women to state their current weight. Assume the mean weight in the sample was ?¯=161.

We will treat these data as an SRS from a normally distributed population with standard deviation ?=36 pounds.

Give a 99% confidence interval for the mean weight of adult women based on these data. Enter the upper and lower values of your confidence interval into the spaces provided rounded to two decimal places.

lower value = pounds

upper value = pounds

Do you trust the interval you computed as a 99% confidence interval for the mean weight of all U.S. adult women? Select an answer choice that correctly explains why or why not.

This interval can be trusted since a 99% confidence interval can be expected to be relatively accurate.

This interval would probably be more trustworthy if the poll contacted women with a wider range of different weights.

This interval can be trusted and seen to be around 99% accurate.

There is probably little reason to trust this interval; it is possible that many of the women either wouldn’t know their current weight or would lie about it.

Answer 1

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.5758   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   36.0000   / √   491   =   1.6247
margin of error, E=Z*SE =   2.5758   *   1.6247   =   4.1848
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    161.00   -   4.184840   =   156.8152
Interval Upper Limit = x̅ + E =    161.00   -   4.184840   =   165.1848
99%   confidence interval is (   156.82   < µ <   165.18   )
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There is probably little reason to trust this interval; it is possible that many of the women either wouldn’t know their current weight or would lie about it.