To test the effect of a physical fitness course on one's physical ability, the number of sit-ups that a person could do in one minute, both before and after the course, was recorded. Ten individuals are randomly selected to participate in the course. The results are displayed in the following table. Using this data, find the $99 \%$ confidence interval for the true difference in the number of sit-ups each person can do before and after the course. Assume that the numbers of sit-ups are normally distributed for the population both before and after completing the course.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline Sit-ups before & 46 & 53 & 36 & 48 & 40 & 33 & 47 & 37 & 47 & 24 \\
\hline Sit-ups after & 55 & 59 & 40 & 57 & 56 & 49 & 60 & 39 & 56 & 26 \\
\hline
\end{tabular}
Step 3 of 4: Calculate the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.
Answer (2)
Calculate the difference in sit-ups for each individual.
Find the mean d ˉ and standard deviation s d of these differences.
Determine the critical t-value t α /2 , n − 1 for a 99% confidence level with 9 degrees of freedom.
Compute the margin of error: E = t α /2 , n − 1 ⋅ n s d = 5.355385 .
Explanation
Understand the problem and provided data We are given paired data of sit-ups before and after a physical fitness course for 10 individuals. Our goal is to find the 99% confidence interval for the true difference in the number of sit-ups and then calculate the margin of error. The sample size is n = 10, and the confidence level is 99%.
Calculate the differences First, we need to calculate the difference in sit-ups for each individual (After - Before). The differences are: 55-46=9, 59-53=6, 40-36=4, 57-48=9, 56-40=16, 49-33=16, 60-47=13, 39-37=2, 56-47=9, 26-24=2.
Calculate the mean of the differences Next, we calculate the sample mean of the differences, denoted as d ˉ .
d ˉ = 10 9 + 6 + 4 + 9 + 16 + 16 + 13 + 2 + 9 + 2 = 10 86 = 8.6
Calculate the standard deviation of the differences Now, we calculate the sample standard deviation of the differences, denoted as s d .
First, we calculate the squared differences from the mean: ( 9 − 8.6 ) 2 = 0.16 ( 6 − 8.6 ) 2 = 6.76 ( 4 − 8.6 ) 2 = 21.16 ( 9 − 8.6 ) 2 = 0.16 ( 16 − 8.6 ) 2 = 54.76 ( 16 − 8.6 ) 2 = 54.76 ( 13 − 8.6 ) 2 = 19.36 ( 2 − 8.6 ) 2 = 43.56 ( 9 − 8.6 ) 2 = 0.16 ( 2 − 8.6 ) 2 = 43.56
Sum of squared differences = 0.16 + 6.76 + 21.16 + 0.16 + 54.76 + 54.76 + 19.36 + 43.56 + 0.16 + 43.56 = 244.4
Then, we calculate the sample standard deviation: s d = n − 1 ∑ i = 1 n ( d i − d ˉ ) 2 = 10 − 1 244.4 = 9 244.4 = 27.155556 ≈ 5.211099
Find the critical t-value We need to find the critical value, t α /2 , n − 1 , for a 99% confidence level with n − 1 = 9 degrees of freedom. Since it is a 99% confidence interval, α = 1 − 0.99 = 0.01 , so α /2 = 0.005 . We need to find t 0.005 , 9 . From the t-distribution table or calculator, t 0.005 , 9 ≈ 3.249836 .
Calculate the margin of error Now, we calculate the margin of error using the formula: E = t α /2 , n − 1 ⋅ n s d = 3.249836 ⋅ 10 5.211099 ≈ 3.249836 ⋅ 3.162278 5.211099 ≈ 3.249836 ⋅ 1.647835 ≈ 5.355385
Round the margin of error Finally, we round the margin of error to six decimal places: 5.355385.
State the final answer The margin of error for the 99% confidence interval for the true difference in the number of sit-ups each person can do before and after the course is approximately 5.355385.
Examples
Understanding the margin of error is crucial in various fields. For instance, in medical research, when testing a new drug's effectiveness, the margin of error helps determine the reliability of the results. If a fitness course claims to improve sit-up performance, calculating the confidence interval and margin of error can validate whether the improvement is statistically significant or merely due to chance. This ensures that claims are evidence-based and reliable.
The margin of error for the 99% confidence interval regarding the true difference in the number of sit-ups before and after the fitness course is approximately 5.355385. This calculation involved determining the differences, calculating the mean and standard deviation of those differences, and utilizing a critical t-value for a 99% confidence level.
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