Question

751 body temperature measurements were taken. The sample data resulted in a sample mean of 98.1 F and a sample standard deviation of 0.7 F. Use the traditional method and a 0.05 significance level to test the claim that the mean body temperature is less than 98.6 F.

Answer 1

There is significant evidence to reject the null hypothesis that the mean body temperature is 98.6 F. The sample data suggests a lower average body temperature.

Step-by-step explanation:

1. Define the null and alternative hypotheses:

Null hypothesis (H0): The mean body temperature is 98.6 F (μ = 98.6)

Alternative hypothesis (Ha): The mean body temperature is less than 98.6 F (μ < 98.6)

2. Calculate the test statistic:

Z = (Xbar - μ0) / (σ / √n)

Z = (98.1 - 98.6) / (0.7 / √751) ≈ -3.14

3. Determine the critical value at the 0.05 significance level:

Since it's a one-tailed test (testing for less than), the critical Z-score is -1.645.

4. Compare the test statistic and critical value:

The calculated Z-score (-3.14) is more extreme than the critical value (-1.645).

5. Make a decision:

Reject the null hypothesis (H0). There is significant evidence to conclude that the mean body temperature is less than 98.6 F.

In simpler terms: The sample data shows a statistically significant difference from the claimed average body temperature of 98.6 F. It suggests a lower average body temperature in the population, with a confidence level of 95%.

Answer 2

The mean value of the sample is 98.1 F and its standard deviation is 0.7 F.

The margin of error of the mean value is given by 0.7/sqrt(751) = 0.7/27.4 = 0.026 (rounded to the nearest thousandth)

Using the Z test, we got: Z = (98.6 - 98.1)/(0.026) = 0.5/0.026 = 196

Therefore, the mean value of the sample is incompatible with 98.6 and we can claim that the mean body temperature is less than it.