Quiz 6

M 154a, Nov. 21, 1997
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It is commonly believed that "normal'' body temperature is 98.7 degrees Farenheit. Of course, this varies slightly from individual to individual. For the sake of this problem, let's suppose that the body temperature of a randomly selected person, call it X, follows a normal distribution with variance equal to 1.5 degrees. Suppose a researcher measures the temperature of 10 people with the following results:

97.4, 100.5, 98.4, 98.4, 99.1, 99.6,  99.5, 100.3, 98.5, 97.4

1.

Find a 90% confidence interval for the mean body temperature based on these data. Solution The standard deviation is known and is the squareroot of 1.5. So a CI is ${\bar X} \pm z_{\alpha/2}{\sigma \over {\sqrt n}}$ which is $98.91 \pm 1.64 {\sqrt .151}$ = (98.27,99.54).

2.

Suppose the variance was unknown. The sample variance of this list, s², is 1.21. Find a 90% confidence interval for the mean body temperature when the variance is unknown.

Solution Now the variance is unknown, but the sample variance estimates it at 1.21. The proper confidence interval is now based on a t distribution with n-1 = 9 degrees of freedom. $98.91 \pm 1.83 {\sqrt .121} = (98.27, 99.55)$. The confidence intervals don't differ very much here (at least to two decimal places) because although the value of the constant in the margin of error increased in the second part (from 1.64 to 1.83), by chance the estimate of the standard error decreased enough to make up for it.