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A popular radio show recently suggested that when commuting on a bicycle, people tend to eat more doughnuts. To test this, you found a group of 8 people who would be willing to ride a bike to work for the next month. You first figure out how many doughnuts per week they ate on a normal basis, then you have them commute to work on a bike for a whole month. During the last week of their biking, you again measure how many doughnuts they eat (the data is provided below).

Determine if there is any truth to the claims made in the radio show. Assume an alpha level of 0.05.

Doughnuts Eaten:

- Before Bike Commuting: 2, 3, 6, 7, 4, 8, 6, 4
- After Bike Commuting: 4, 5, 7, 8, 10, 8, 8, 6

Note: Please show all of the steps we covered when formally testing hypotheses.

Answer :

From the data and the results of the paired t-test, we have evidence to support the claim made in the radio show that people tend to eat more doughnuts after bike commuting

How do we calculate?

The Null hypothesis states that there is no difference in the average number of doughnuts eaten before and after bike commuting.

The alternative hypothesis states that here is a difference in the average number of doughnuts eaten before and after bike commuting.

Differences = After Bike Commuting - Before Bike Commuting

Differences = (4-2), (5-3), (7-6), (8-7), (10-4), (8-8), (8-6), (6-4)

Differences = 2, 2, 1, 1, 6, 0, 2, 2

Sample mean= sum of differences / number of differences

Sample mean = ( (2 + 2 + 1 + 1 + 6 + 0 + 2 + 2) / 8

Sample mean = 2

Sample standard deviation (s) =√ [(sum of (difference - sample mean)²) / (number of differences - 1)]

Sample standard deviation (s) = √[(0² + 0²+ (-1)² + (-1)² + (4)² + (-2)² + 0² + 0²) / (8 - 1)]

Sample standard deviation (s) = √(10/7)

Sample standard deviation (s) = 1.195

t = (sample mean - μ) / (s / √n)

t = (2 - 0) / (1.195 / √8)

t = 2 / (1.195 / 2.828)

t = 2 / 0.423

t = 4.724

The critical t-value is obtained from a t-distribution table to be 2.365.

we can reject the null hypothesis, because the absolute value of the calculated t-value is greater than the critical t-value.

Learn more about t-value at:

https://brainly.com/question/27192813

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