To estimate the true mean price for the camera with 95% confidence, we can follow these steps:
- Calculate the sample mean (\(\bar{x}\)) and the sample standard deviation (s).
- Determine the t-score for 95% confidence with n-1 degrees of freedom.
- Calculate the margin of error.
- 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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