Question
The prices in dollars for a particular model of camera are shown below for 10 online retailers. Estimate the true mean price for this particular camera with 95% confidence. Assume normal distribution.
225, 240, 215, 206, 211, 210, 193, 250, 225, 202
Please round to the nearest tenth and write your answer in the form ( , ); for example (31,34) or (25,46). Do not type in any extra spaces.
225, 240, 215, 206, 211, 210, 193, 250, 225, 202
Please round to the nearest tenth and write your answer in the form ( , ); for example (31,34) or (25,46). Do not type in any extra spaces.
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Answered by
GPT-4o mini
To estimate the true mean price for the camera with 95% confidence, we can follow these steps:
1. Calculate the sample mean (\(\bar{x}\)) and the sample standard deviation (s).
2. Determine the t-score for 95% confidence with n-1 degrees of freedom.
3. Calculate the margin of error.
4. Construct the confidence interval.
**Step 1: Calculate the sample mean and standard deviation**
Given prices: 225, 240, 215, 206, 211, 210, 193, 250, 225, 202
- Sample size (n) = 10
Calculating the sample mean (\(\bar{x}\)):
\[
\bar{x} = \frac{(225 + 240 + 215 + 206 + 211 + 210 + 193 + 250 + 225 + 202)}{10} = \frac{ 2250}{10} = 225.0
\]
Calculating the sample standard deviation (s):
\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
\]
Calculating each deviation from the mean:
- \((225 - 225)^2 = 0\)
- \((240 - 225)^2 = 225\)
- \((215 - 225)^2 = 100\)
- \((206 - 225)^2 = 361\)
- \((211 - 225)^2 = 196\)
- \((210 - 225)^2 = 225\)
- \((193 - 225)^2 = 1024\)
- \((250 - 225)^2 = 625\)
- \((225 - 225)^2 = 0\)
- \((202 - 225)^2 = 529\)
Sum of squared deviations:
\[
0 + 225 + 100 + 361 + 196 + 225 + 1024 + 625 + 0 + 529 = 3285
\]
Now calculate s:
\[
s = \sqrt{\frac{3285}{10 - 1}} = \sqrt{\frac{3285}{9}} \approx \sqrt{364.33} \approx 19.1
\]
**Step 2: Determine the t-score for 95% confidence**
Degrees of freedom (df) = n - 1 = 10 - 1 = 9.
Using a t-table, the t-score for 9 degrees of freedom at 95% confidence is approximately equal to 2.262.
**Step 3: Calculate the margin of error**
Margin of error (ME):
\[
ME = t \cdot \frac{s}{\sqrt{n}} = 2.262 \cdot \frac{19.1}{\sqrt{10}} \approx 2.262 \cdot 6.032 = 13.66 \approx 13.7
\]
**Step 4: Construct the confidence interval**
The confidence interval is given by:
\[
(\bar{x} - ME, \bar{x} + ME) = (225.0 - 13.7, 225.0 + 13.7) = (211.3, 238.7)
\]
Therefore, the 95% confidence interval for the true mean price of the camera is:
\((211.3, 238.7)\)
1. Calculate the sample mean (\(\bar{x}\)) and the sample standard deviation (s).
2. Determine the t-score for 95% confidence with n-1 degrees of freedom.
3. Calculate the margin of error.
4. Construct the confidence interval.
**Step 1: Calculate the sample mean and standard deviation**
Given prices: 225, 240, 215, 206, 211, 210, 193, 250, 225, 202
- Sample size (n) = 10
Calculating the sample mean (\(\bar{x}\)):
\[
\bar{x} = \frac{(225 + 240 + 215 + 206 + 211 + 210 + 193 + 250 + 225 + 202)}{10} = \frac{ 2250}{10} = 225.0
\]
Calculating the sample standard deviation (s):
\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
\]
Calculating each deviation from the mean:
- \((225 - 225)^2 = 0\)
- \((240 - 225)^2 = 225\)
- \((215 - 225)^2 = 100\)
- \((206 - 225)^2 = 361\)
- \((211 - 225)^2 = 196\)
- \((210 - 225)^2 = 225\)
- \((193 - 225)^2 = 1024\)
- \((250 - 225)^2 = 625\)
- \((225 - 225)^2 = 0\)
- \((202 - 225)^2 = 529\)
Sum of squared deviations:
\[
0 + 225 + 100 + 361 + 196 + 225 + 1024 + 625 + 0 + 529 = 3285
\]
Now calculate s:
\[
s = \sqrt{\frac{3285}{10 - 1}} = \sqrt{\frac{3285}{9}} \approx \sqrt{364.33} \approx 19.1
\]
**Step 2: Determine the t-score for 95% confidence**
Degrees of freedom (df) = n - 1 = 10 - 1 = 9.
Using a t-table, the t-score for 9 degrees of freedom at 95% confidence is approximately equal to 2.262.
**Step 3: Calculate the margin of error**
Margin of error (ME):
\[
ME = t \cdot \frac{s}{\sqrt{n}} = 2.262 \cdot \frac{19.1}{\sqrt{10}} \approx 2.262 \cdot 6.032 = 13.66 \approx 13.7
\]
**Step 4: Construct the confidence interval**
The confidence interval is given by:
\[
(\bar{x} - ME, \bar{x} + ME) = (225.0 - 13.7, 225.0 + 13.7) = (211.3, 238.7)
\]
Therefore, the 95% confidence interval for the true mean price of the camera is:
\((211.3, 238.7)\)
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