Body temperatures in degrees Fahrenheit have been recorded f

Body temperatures (in degrees Fahrenheit) have been recorded for a sample of 130 healthy adults (Shoemaker, 1996). The sample mean body temperature is 98.249°F, and the sample standard deviation is 0.733°F.

d. Calculate a 95% confidence interval for the population mean body temperature, based on the sample results for these 130 healthy adults. [Round your answer to 2 decimal places.]

g. Suppose the sample size had been 13 rather than 130, but the sample statistics turned out exactly the same. How would you expect a 95% confidence interval to differ in this case from the 95% interval in part d?

h. Produce the confidence interval mentioned in part g, and describe how the interval has changed. [Round your answer to 1 decimal place.]

95% CI is?

Midpoint is?

width is?

Solution

d)

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 =    0.025          
X = sample mean =    98.249          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    0.733          
n = sample size =    130          
              
Thus,              
Margin of Error E =    0.126002856          
Lower bound =    98.12299714          
Upper bound =    98.37500286          
              
Thus, the confidence interval is              
              
(   98.12299714   ,   98.37500286   ) [ANSWER]

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g)

This would produce a wider confidence interval as the standard error would be larger. You would also use t instead of z, which is always larger as well.

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h)

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    98.249          
t(alpha/2) = critical t for the confidence interval =    2.17881283          
s = sample standard deviation =    0.733          
n = sample size =    13          
df = n - 1 =    12          
Thus,              
Margin of Error E =    0.442947467          
Lower bound =    97.80605253          
Upper bound =    98.69194747          
              
Thus, the confidence interval is              
              
(   97.80605253   ,   98.69194747   ) [ANSWER]

Midpoint = X = 98.429 [ANSWER]

width = 2*E = 0.885894934 [ANSWER]

So, it is now wider, but it still has the same midpoint.