Answer :
a) The value of the test statistic is approximately -1.604. b) The 1% critical value for this test is approximately -2.920.
(a) The test statistic to determine if the mean body temperature is less than 98.6 °F can be calculated using the sample data. We need to perform a one-sample t-test using the formula:
t = (x - μ) / (s / √n)
where x is the sample mean, μ is the hypothesized population mean (98.6 °F), s is the sample standard deviation, and n is the sample size.
Using the given sample data, we have:
x = (98.2 + 98.2 + 98.8 + 98.4 + 97.4 + 98.2 + 98.0 + 97.5) / 8 = 98.25
s = √((98.2 - 98.25)² + (98.2 - 98.25)² + (98.8 - 98.25)² + (98.4 - 98.25)² + (97.4 - 98.25)² + (98.2 - 98.25)² + (98.0 - 98.25)² + (97.5 - 98.25)² / 7) = 0.559
n = 8
Plugging these values into the formula, we get:
t = (98.25 - 98.6) / (0.559 / √8) ≈ -1.604
Therefore, the value of the test statistic is approximately -1.604.
(b) To find the 1% critical value for the one-sample t-test, we need to consult the t-distribution table or use statistical software. The critical value corresponds to the cutoff point below which we reject the null hypothesis.
Since we are conducting a one-tailed test (we want to determine if the mean body temperature is less than 98.6 °F), we will look for the critical value in the left tail of the t-distribution.
The degrees of freedom for our test is n - 1, which in this case is 8 - 1 = 7. With a significance level of 1%, we need to find the t-value that has a cumulative probability of 0.01 in the left tail of the t-distribution with 7 degrees of freedom.
Using a t-distribution table or statistical software, we find that the 1% critical value for a one-tailed t-test with 7 degrees of freedom is approximately -2.920.
Therefore, the 1% critical value for this test is approximately -2.920.
Learn more about sample mean here:
brainly.com/question/33323852
#SPJ11
