Question
Breathing rates, in breaths per minute, were measured for a group of ten subjects at rest, and then during moderate exercise. The results were as follows:
Subject 1: rest- 20 , exercise 40
2: 19/35
3: 17/37
4: 15/39
5: 14/39
6: 18/32
7: 20/35
8: 21/31
9: 18/35
10: 16/37
Construct a 99% confidence interval for the mean increase in breathing rate due to exercise. Let d represent the breathing rate after exercise minus the breathing rate at rest. Round to 1 decimal place
Subject 1: rest- 20 , exercise 40
2: 19/35
3: 17/37
4: 15/39
5: 14/39
6: 18/32
7: 20/35
8: 21/31
9: 18/35
10: 16/37
Construct a 99% confidence interval for the mean increase in breathing rate due to exercise. Let d represent the breathing rate after exercise minus the breathing rate at rest. Round to 1 decimal place
Answers
GPT-4o mini
To construct a 99% confidence interval for the mean increase in breathing rate due to exercise, we first need to calculate the differences \( d \) for each subject (exercise breathing rate - rest breathing rate), find the mean and standard deviation of these differences, and then use that to construct the confidence interval.
### Step 1: Calculate the differences \( d \)
Here are the differences for each subject:
1. Subject 1: \( d_1 = 40 - 20 = 20 \)
2. Subject 2: \( d_2 = 35 - 19 = 16 \)
3. Subject 3: \( d_3 = 37 - 17 = 20 \)
4. Subject 4: \( d_4 = 39 - 15 = 24 \)
5. Subject 5: \( d_5 = 39 - 14 = 25 \)
6. Subject 6: \( d_6 = 32 - 18 = 14 \)
7. Subject 7: \( d_7 = 35 - 20 = 15 \)
8. Subject 8: \( d_8 = 31 - 21 = 10 \)
9. Subject 9: \( d_9 = 35 - 18 = 17 \)
10. Subject 10: \( d_{10} = 37 - 16 = 21 \)
Now, we have the differences:
- \( d = [20, 16, 20, 24, 25, 14, 15, 10, 17, 21] \)
### Step 2: Calculate the sample mean \( \bar{d} \)
\[
\bar{d} = \frac{20 + 16 + 20 + 24 + 25 + 14 + 15 + 10 + 17 + 21}{10} = \frac{ 20 + 16 + 20 + 24 + 25 + 14 + 15 + 10 + 17 + 21}{10} = \frac{ 192 }{ 10} = 19.2
\]
### Step 3: Calculate the sample standard deviation \( s_d \)
To find the sample standard deviation, we first calculate the variance.
1. Compute each \( (d_i - \bar{d})^2 \):
\[
\begin{align*}
(d_1 - \bar{d})^2 & = (20 - 19.2)^2 = 0.64 \\
(d_2 - \bar{d})^2 & = (16 - 19.2)^2 = 10.24 \\
(d_3 - \bar{d})^2 & = (20 - 19.2)^2 = 0.64 \\
(d_4 - \bar{d})^2 & = (24 - 19.2)^2 = 22.09 \\
(d_5 - \bar{d})^2 & = (25 - 19.2)^2 = 33.64 \\
(d_6 - \bar{d})^2 & = (14 - 19.2)^2 = 27.04 \\
(d_7 - \bar{d})^2 & = (15 - 19.2)^2 = 17.64 \\
(d_8 - \bar{d})^2 & = (10 - 19.2)^2 = 84.64 \\
(d_9 - \bar{d})^2 & = (17 - 19.2)^2 = 4.84 \\
(d_{10} - \bar{d})^2 & = (21 - 19.2)^2 = 3.24 \\
\end{align*}
\]
Now summing these:
\[
\sum (d_i - \bar{d})^2 = 0.64 + 10.24 + 0.64 + 22.09 + 33.64 + 27.04 + 17.64 + 84.64 + 4.84 + 3.24 = 195.1
\]
Now, calculate the variance:
\[
s_d^2 = \frac{\sum (d_i - \bar{d})^2}{n - 1} = \frac{195.1}{10 - 1} = \frac{195.1}{9} \approx 21.67
\]
Thus, the standard deviation \( s_d \):
\[
s_d = \sqrt{21.67} \approx 4.65
\]
### Step 4: Construct the 99% confidence interval
Using the t-distribution for \( n - 1 = 9 \) degrees of freedom and a 99% confidence level, we look up the critical value of \( t \). For \( df = 9 \), the critical t-value \( t_{0.005, 9} \approx 3.25 \) (you can find this value in t-distribution tables).
\[
\text{Margin of error} = t \cdot \left(\frac{s_d}{\sqrt{n}}\right) = 3.25 \cdot \left(\frac{4.65}{\sqrt{10}}\right)
\]
Calculating \( \frac{s_d}{\sqrt{n}} \):
\[
\frac{4.65}{\sqrt{10}} \approx \frac{4.65}{3.16} \approx 1.47
\]
Now, calculating the margin of error:
\[
\text{Margin of error} = 3.25 \cdot 1.47 \approx 4.78
\]
### Final Confidence Interval
Now, calculate the 99% confidence interval:
\[
\text{CI} = \bar{d} \pm \text{Margin of error} = 19.2 \pm 4.78
\]
Thus, the interval is:
\[
(19.2 - 4.78, 19.2 + 4.78) = (14.42, 23.98)
\]
Rounding to one decimal place, the final confidence interval for the mean increase in breathing rate due to exercise is:
\[
\boxed{(14.4, 24.0)}
\]
### Step 1: Calculate the differences \( d \)
Here are the differences for each subject:
1. Subject 1: \( d_1 = 40 - 20 = 20 \)
2. Subject 2: \( d_2 = 35 - 19 = 16 \)
3. Subject 3: \( d_3 = 37 - 17 = 20 \)
4. Subject 4: \( d_4 = 39 - 15 = 24 \)
5. Subject 5: \( d_5 = 39 - 14 = 25 \)
6. Subject 6: \( d_6 = 32 - 18 = 14 \)
7. Subject 7: \( d_7 = 35 - 20 = 15 \)
8. Subject 8: \( d_8 = 31 - 21 = 10 \)
9. Subject 9: \( d_9 = 35 - 18 = 17 \)
10. Subject 10: \( d_{10} = 37 - 16 = 21 \)
Now, we have the differences:
- \( d = [20, 16, 20, 24, 25, 14, 15, 10, 17, 21] \)
### Step 2: Calculate the sample mean \( \bar{d} \)
\[
\bar{d} = \frac{20 + 16 + 20 + 24 + 25 + 14 + 15 + 10 + 17 + 21}{10} = \frac{ 20 + 16 + 20 + 24 + 25 + 14 + 15 + 10 + 17 + 21}{10} = \frac{ 192 }{ 10} = 19.2
\]
### Step 3: Calculate the sample standard deviation \( s_d \)
To find the sample standard deviation, we first calculate the variance.
1. Compute each \( (d_i - \bar{d})^2 \):
\[
\begin{align*}
(d_1 - \bar{d})^2 & = (20 - 19.2)^2 = 0.64 \\
(d_2 - \bar{d})^2 & = (16 - 19.2)^2 = 10.24 \\
(d_3 - \bar{d})^2 & = (20 - 19.2)^2 = 0.64 \\
(d_4 - \bar{d})^2 & = (24 - 19.2)^2 = 22.09 \\
(d_5 - \bar{d})^2 & = (25 - 19.2)^2 = 33.64 \\
(d_6 - \bar{d})^2 & = (14 - 19.2)^2 = 27.04 \\
(d_7 - \bar{d})^2 & = (15 - 19.2)^2 = 17.64 \\
(d_8 - \bar{d})^2 & = (10 - 19.2)^2 = 84.64 \\
(d_9 - \bar{d})^2 & = (17 - 19.2)^2 = 4.84 \\
(d_{10} - \bar{d})^2 & = (21 - 19.2)^2 = 3.24 \\
\end{align*}
\]
Now summing these:
\[
\sum (d_i - \bar{d})^2 = 0.64 + 10.24 + 0.64 + 22.09 + 33.64 + 27.04 + 17.64 + 84.64 + 4.84 + 3.24 = 195.1
\]
Now, calculate the variance:
\[
s_d^2 = \frac{\sum (d_i - \bar{d})^2}{n - 1} = \frac{195.1}{10 - 1} = \frac{195.1}{9} \approx 21.67
\]
Thus, the standard deviation \( s_d \):
\[
s_d = \sqrt{21.67} \approx 4.65
\]
### Step 4: Construct the 99% confidence interval
Using the t-distribution for \( n - 1 = 9 \) degrees of freedom and a 99% confidence level, we look up the critical value of \( t \). For \( df = 9 \), the critical t-value \( t_{0.005, 9} \approx 3.25 \) (you can find this value in t-distribution tables).
\[
\text{Margin of error} = t \cdot \left(\frac{s_d}{\sqrt{n}}\right) = 3.25 \cdot \left(\frac{4.65}{\sqrt{10}}\right)
\]
Calculating \( \frac{s_d}{\sqrt{n}} \):
\[
\frac{4.65}{\sqrt{10}} \approx \frac{4.65}{3.16} \approx 1.47
\]
Now, calculating the margin of error:
\[
\text{Margin of error} = 3.25 \cdot 1.47 \approx 4.78
\]
### Final Confidence Interval
Now, calculate the 99% confidence interval:
\[
\text{CI} = \bar{d} \pm \text{Margin of error} = 19.2 \pm 4.78
\]
Thus, the interval is:
\[
(19.2 - 4.78, 19.2 + 4.78) = (14.42, 23.98)
\]
Rounding to one decimal place, the final confidence interval for the mean increase in breathing rate due to exercise is:
\[
\boxed{(14.4, 24.0)}
\]