Question

Question 1:

A small data set is given below. You will run a t-test using Excel. In order to receive full credit, you must turn in Excel output and your final answers. Round off final answers to three decimal places, if appropriate.

A medical researcher wants to determine whether a drug changes the body’s temperature. Seventest subjects are randomly selected, and the body temperature (in degrees Fahrenheit) of each is measured. The subjects are then given the drug and after 20 minutes, the body temperature of each is measured again. The results are listed below. Use significance level α = 0.05, and we assume that the body temperatures are normally distributed.

Subject	1	2	3	4	5	6	7
Initial Temperature	101.8	98.5	98.1	99.4	98.9	100.2	97.9
Second Temperature	99.2	98.4	98.2	99.0	98.6	99.7	97.8

Provide results for the following questions:

1. Which t-test you used? (2 p)

2. Find the degrees of freedom? (2 p)

3. What is the t-statistic of the test? (3 p)

4. Find the t-critical value(s)? (3 p)

5. One or two tailed hypothesis testing used? (5p)

6. What is the corresponding p value? (5 p)

7. Do you reject or fail to reject the null hypothesis? (5 p)

8. Is there enough evidence to conclude that the drug changes the body’s temperature? (α =

   0.05). What evidence do you use to support your conclusion? (10 p)

PLEASE SHOW WORK THANK YOU!

Answer 1

By using Excel we have to solve this question.

Enter data into excel.

The null and alternative hypothesis is

Level of significance = 0.05

Which t-test you used?

Data is dependent so we have to use matched paired t-test.

Find the degrees of freedom?

Sample size = n = 7

Degrees of freedom = n - 1 = 7 - 1 = 6

In excel click on Data ------> Data Analysis ------>t-Test: Paired Two Sample for Means----->Ok

Variable 1 Range: Select initial values

Variable 2 Range:Select Second values

Hypothesized Mean Difference: 0

Alpha: 0.05

Output Range: Select any empty cell

---->Ok

We get

Test statistic = t = 1.5962

The t-critical value = 2.4469

Two-tailed test.

The p-value = 0.1616

P-value > 0.05 we fail to reject null hypothesis.

There is not enough evidence to conclude that the drug changes the body’s temperature.

Evidence: P-value > 0.05 we fail to reject null hypothesis.