High School

The resting body temperatures of 25 test subjects (in °F) were found to be:
97.8, 97.2, 97.4, 97.6, 97.8, 97.9, 98.0, 98.0, 98.0, 98.1, 98.2, 98.3, 98.3, 98.4, 98.4, 98.4, 98.5, 98.6, 98.6, 98.7, 98.8, 98.8, 98.9, 98.9, 99.0

a. Test the hypothesis \( H_0: \mu = 98.6 \) versus the alternative \( H_a: \mu \neq 98.6 \). Use \( \alpha = 0.05 \).

b. Construct a 95% confidence interval for the mean resting body temperature.

Answer :

a. The mean resting body temperature is 98.6°F, which is the null hypothesis. We will use a two-tailed t-test with a significance level of 0.05. b. The 95% confidence interval for the mean resting body temperature is (98.07°F, 98.61°F).


a. To test the hypothesis, we compare the mean resting body temperature (sample mean) with the assumed population mean of 98.6°F.

We calculate the t-value using the formula:

t = (sample mean - assumed population mean) / (sample standard deviation / √n).

With a two-tailed test, we divide the significance level (0.05) by 2 and find the corresponding critical t-values. We reject the null hypothesis if the estimated t-value exceeds the crucial t-values.

b. To construct a 95% confidence interval, we use the formula:

CI = sample mean ± (critical value * standard error).

The critical value can be found using the t-distribution table with the degrees of freedom (n-1) and a confidence level of 95%. The standard error is calculated as the sample standard deviation divided by the square root of the sample size. We then calculate the lower and upper limits of the confidence interval using these values.

Learn more about null hypothesis here:

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