A software developer wants to know how many new computer games people buy each year. A sample of 271 people was taken to study their purchasing habits. Construct the $90 \%$ confidence interval for the mean number of computer games purchased each year if the sample mean was found to be 6.3. Assume that the population standard deviation is 1.2. Round your answers to one decimat place.
Answer How to enter your answer (opens in new window)
2 Points
Tables
Keypad
Keyboard Shortcuts
Lower endpoint:
Upper endpoint
Prev
Next

Step 1 ：Given values are: sample mean (\(\bar{x}\)) = 6.3, population standard deviation (\(\sigma\)) = 1.2, sample size (n) = 271, and Z-score for 90% confidence level = 1.645.

Step 2 ：First, calculate the margin of error using the formula: Margin of Error = Z-score * \(\sigma / \sqrt{n}\).

Step 3 ：Substitute the given values into the formula: Margin of Error = 1.645 * (1.2 / \(\sqrt{271}\)) = 0.11991196032415735.

Step 4 ：Next, calculate the confidence interval using the formula: Confidence Interval = \(\bar{x} \pm\) Margin of Error.

Step 5 ：Substitute the values into the formula: Lower endpoint = 6.3 - 0.11991196032415735 = 6.2 (rounded to one decimal place), Upper endpoint = 6.3 + 0.11991196032415735 = 6.4 (rounded to one decimal place).

Step 6 ：Final Answer: The 90% confidence interval for the mean number of computer games purchased each year is \(\boxed{[6.2, 6.4]}\).