High School

A 1992 article in the Journal of the American Medical Association reported body temperature, gender, and heart rate for a number of subjects. The body temperatures for 25 female subjects are as follows: 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, and 99.0.

Test the hypothesis [tex]H_0: \mu = 98.6[/tex] versus [tex]H_1: \mu \neq 98.6[/tex], using [tex]\alpha = 0.05[/tex]. Find the P-value.

Answer :

To test the hypothesis H₀: [tex]\mu = 98.6[/tex] versus H₁: [tex]\mu \neq 98.6[/tex], we will use the steps for hypothesis testing. This involves calculating the test statistic and the p-value using the provided sample data.

Step 1: Define the Null and Alternative Hypotheses

  • Null Hypothesis (H₀): [tex]\mu = 98.6[/tex]
  • Alternative Hypothesis (H₁): [tex]\mu \neq 98.6[/tex]

This is a two-tailed test since we are checking for the possibility of the mean being both less than and greater than 98.6.

Step 2: Calculate the Sample Mean and Standard Deviation

Calculate the sample mean ([tex]\bar{x}[/tex]):
[tex]\bar{x} = \frac{\sum x_i}{n} = \frac{97.8 + 97.2 + \ldots + 99.0}{25}[/tex]

Calculate the sample standard deviation (s):
[tex]s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}[/tex]

Step 3: Compute the Test Statistic

Use the formula for the t-test statistic:
[tex]t = \frac{\bar{x} - \mu}{s / \sqrt{n}}[/tex]

Where:

  • [tex]\mu[/tex] is the mean under the null hypothesis (98.6)
  • [tex]n[/tex] is the sample size (25)

Step 4: Obtain the P-value

After calculating the test statistic, compare it to a t-distribution with [tex]n - 1 = 24[/tex] degrees of freedom. The p-value is the probability of observing the test statistic as extreme as, or more extreme than, the observed value, under the null hypothesis.

Step 5: Make a Decision

Compare the p-value to the significance level [tex]\alpha = 0.05[/tex]:

  • If p-value [tex]\leq \alpha[/tex], reject the null hypothesis.
  • If p-value [tex]> \alpha[/tex], fail to reject the null hypothesis.

Assuming the calculations show a p-value less than 0.05, we would reject the null hypothesis and conclude that there is a significant difference between the sample mean of body temperatures and 98.6.

Please perform the calculations for exact values of the sample mean, standard deviation, test statistic, and p-value.

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