Question
The reaction times for a random sample of 8 persons to a stimulant were recorded as:
3
4
3.1
5
2.9
2.3
3
4.1
Find a 90% confidence interval for the mean of all reaction times. (Normality of reaction times is assumed).
3
4
3.1
5
2.9
2.3
3
4.1
Find a 90% confidence interval for the mean of all reaction times. (Normality of reaction times is assumed).
Answers
GPT-4o mini
To find a 90% confidence interval for the mean reaction times based on the sample, we'll follow these steps:
1. **Calculate the sample mean (\(\bar{x}\))**.
2. **Calculate the sample standard deviation (\(s\))**.
3. **Determine the critical t-value for a 90% confidence interval**.
4. **Calculate the margin of error**.
5. **Construct the confidence interval**.
### Step 1: Calculate the Sample Mean (\(\bar{x}\))
The reaction times are:
- 3.0, 4.0, 3.1, 5.0, 2.9, 2.3, 3.0, 4.1
\[
\bar{x} = \frac{\sum x}{n} = \frac{3.0 + 4.0 + 3.1 + 5.0 + 2.9 + 2.3 + 3.0 + 4.1}{8}
\]
\[
\bar{x} = \frac{27.4}{8} = 3.425
\]
### Step 2: Calculate the Sample Standard Deviation (\(s\))
Calculate the deviations from the mean, square them, sum them, and then compute \(s\):
\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
\]
Calculating each deviation:
- \( (3.0 - 3.425)^2 = 0.180625 \)
- \( (4.0 - 3.425)^2 = 0.330625 \)
- \( (3.1 - 3.425)^2 = 0.105625 \)
- \( (5.0 - 3.425)^2 = 2.505625 \)
- \( (2.9 - 3.425)^2 = 0.275625 \)
- \( (2.3 - 3.425)^2 = 1.265625 \)
- \( (3.0 - 3.425)^2 = 0.180625 \)
- \( (4.1 - 3.425)^2 = 0.455625 \)
Summing them up:
\[
\sum (x_i - \bar{x})^2 = 0.180625 + 0.330625 + 0.105625 + 2.505625 + 0.275625 + 1.265625 + 0.180625 + 0.455625 = 5.025
\]
Now calculate \(s\):
\[
s = \sqrt{\frac{5.025}{8 - 1}} = \sqrt{\frac{5.025}{7}} \approx \sqrt{0.717857} \approx 0.847
\]
### Step 3: Determine the Critical t-value
For a 90% confidence interval with \(n - 1 = 8 - 1 = 7\) degrees of freedom, we can use the t-distribution table or calculator to find the critical t-value (\(t_{\alpha/2}\)).
Using a t-table for 7 degrees of freedom at the 0.05 significance level:
\[
t_{0.05, 7} \approx 1.895
\]
### Step 4: Calculate the Margin of Error
The margin of error (ME) is calculated as follows:
\[
\text{ME} = t_{\alpha/2} \times \frac{s}{\sqrt{n}} = 1.895 \times \frac{0.847}{\sqrt{8}} \approx 1.895 \times 0.299 = 0.567
\]
### Step 5: Construct the Confidence Interval
Finally, we can construct the confidence interval as follows:
\[
\text{Confidence Interval} = \left( \bar{x} - \text{ME}, \bar{x} + \text{ME} \right) = \left( 3.425 - 0.567, 3.425 + 0.567 \right)
\]
\[
\text{Confidence Interval} = \left( 2.858, 3.992 \right)
\]
### Conclusion
The 90% confidence interval for the mean of all reaction times is approximately \((2.858, 3.992)\).
1. **Calculate the sample mean (\(\bar{x}\))**.
2. **Calculate the sample standard deviation (\(s\))**.
3. **Determine the critical t-value for a 90% confidence interval**.
4. **Calculate the margin of error**.
5. **Construct the confidence interval**.
### Step 1: Calculate the Sample Mean (\(\bar{x}\))
The reaction times are:
- 3.0, 4.0, 3.1, 5.0, 2.9, 2.3, 3.0, 4.1
\[
\bar{x} = \frac{\sum x}{n} = \frac{3.0 + 4.0 + 3.1 + 5.0 + 2.9 + 2.3 + 3.0 + 4.1}{8}
\]
\[
\bar{x} = \frac{27.4}{8} = 3.425
\]
### Step 2: Calculate the Sample Standard Deviation (\(s\))
Calculate the deviations from the mean, square them, sum them, and then compute \(s\):
\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
\]
Calculating each deviation:
- \( (3.0 - 3.425)^2 = 0.180625 \)
- \( (4.0 - 3.425)^2 = 0.330625 \)
- \( (3.1 - 3.425)^2 = 0.105625 \)
- \( (5.0 - 3.425)^2 = 2.505625 \)
- \( (2.9 - 3.425)^2 = 0.275625 \)
- \( (2.3 - 3.425)^2 = 1.265625 \)
- \( (3.0 - 3.425)^2 = 0.180625 \)
- \( (4.1 - 3.425)^2 = 0.455625 \)
Summing them up:
\[
\sum (x_i - \bar{x})^2 = 0.180625 + 0.330625 + 0.105625 + 2.505625 + 0.275625 + 1.265625 + 0.180625 + 0.455625 = 5.025
\]
Now calculate \(s\):
\[
s = \sqrt{\frac{5.025}{8 - 1}} = \sqrt{\frac{5.025}{7}} \approx \sqrt{0.717857} \approx 0.847
\]
### Step 3: Determine the Critical t-value
For a 90% confidence interval with \(n - 1 = 8 - 1 = 7\) degrees of freedom, we can use the t-distribution table or calculator to find the critical t-value (\(t_{\alpha/2}\)).
Using a t-table for 7 degrees of freedom at the 0.05 significance level:
\[
t_{0.05, 7} \approx 1.895
\]
### Step 4: Calculate the Margin of Error
The margin of error (ME) is calculated as follows:
\[
\text{ME} = t_{\alpha/2} \times \frac{s}{\sqrt{n}} = 1.895 \times \frac{0.847}{\sqrt{8}} \approx 1.895 \times 0.299 = 0.567
\]
### Step 5: Construct the Confidence Interval
Finally, we can construct the confidence interval as follows:
\[
\text{Confidence Interval} = \left( \bar{x} - \text{ME}, \bar{x} + \text{ME} \right) = \left( 3.425 - 0.567, 3.425 + 0.567 \right)
\]
\[
\text{Confidence Interval} = \left( 2.858, 3.992 \right)
\]
### Conclusion
The 90% confidence interval for the mean of all reaction times is approximately \((2.858, 3.992)\).