Question

You are on a research team that is investigating the mean temperature of adult humans. The commonly accepted claim is that the mean temperature is about 98.6 degrees F. You want to show that this claim is false. How would you write the null and alternative hypothesis?

Now assume that you have a sample of 35 adults and their mean temperature is 97.8 and the standard deviation of the temperatures is 0.8 Complete the test with alpha = 0.01

1. What is the p-value and why? How did you calculate that?
2. What is the 95% confidence interval based on this sample?
3. Complete the test with critical values and provide the answer ( reject or not)?
4. Complete the test with p-value (reject or not)?
5. Do you reject or not reject the claim, the null or the alternative and why? Is it always like this?
6. What is your conclusion??? Is your conclusion related to the null, the alternative of the claim and why? Is it always like this?

Answer 1

Given that the commonly accepted claim is that the mean temperature is about = 98.6 degrees F.

So, based on the claim the hypotheses are:

Based on the hypothesis it will be a two-tailed test.

assuming that we have a sample of n = 35 adults and their mean temperature is = 97.8 and the standard deviation of the temperatures is s = 0.8.

Since the sample size is greater than 30 and taken from a large population hence it is assumed that it is normally distributed, but since the population, the standard deviation is not known we will be using t-distribution to test the hypothesis.

Rejection region:

Based on the significance level the critical values for the two-tailed test are calculated by excel formula for T-distribution which is =T.INV.2T(0.01,34) , where 30 is the degree of freedom which is calculated as n-1 = 35-1 =34 which results in tc = +/-2.73.

P-value approach:

Reject the Ho if P-value <0.01

So. reject Ho if t-calculated is greater than 2.73 or less than -2.73

Test Statistic:

P-value:

The P-value is again calculated by excel formula for T-distribution which is =T.DIST.2T(5.916,34)

this results in P-value <0.01

Decision:

Since P-value <0.01 and t<-tc hence we reject the null hypothesis.

Conclusion:

Since we can reject the null hypothesis hence we conclude that there is sufficient evidence to warrant the claim.