Answer:
Step-by-step explanation:
Hello!
You need to test the hypothesis that "the average temperature for healthy adults is not equal to 98.6"
The variable of interest is
X: The body temperature of a healthy adult. (Fahrenheit)
And the parameter of interest is the population mean, μ.
The statistic hypotheses are:
H₀: μ = 98.6
H₁: μ ≠ 98.6
α:0.05
Assuming that the variable has a normal distribution you have to use a one-sample t-statistic for this test.
Attached to the answer is the data of body temperature of n=19 healthy adults. Since you didn't copy the raw data for your exercise I'll use this to answer the question.
Using the data the sample mean and standard deviation are:
X[bar]= 98.12
S= 0.69
[tex]t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } ~~t_{n-1}[/tex]
[tex]t_{H_0}= \frac{98.12-98.6}{\frac{0.69}{\sqrt{19} } } = -3.03[/tex]
This test is two-tailed, using the critical value approach, you have the rejection region divided into two tails determined by two critical values:
[tex]t_{n-1;\alpha /2}= t_{18;0.025}= -1.965[/tex]
[tex]t_{n-1;1-\alpha /2}= t_{18;0.975}= 1.965[/tex]
Decision rule:
If [tex]t_{H_0}[/tex] ≤ -1.965 or if [tex]t_{H_0}[/tex] ≥ 1.965, the decision is to reject the null hypothesis.
If -1.965 < [tex]t_{H_0}[/tex] < 1.965, the decision is to not reject the null hypothesis.
The calculated statistic is less than the lower critical value, the decision is to reject the null hypothesis.
Using a significance level of 5%, there is enough evidence to reject the null hypothesis. Then you can conclude that the true mean body temperature for healthy adults is not equal to the traditional 98.6F.
I hope this helps!
