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A student records the repair cost for 17 randomly selected TVs. A sample mean of $73.10 and standard deviation of $16.01 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

2 Answers

4 votes

Final answer:

The 90% confidence interval for the mean repair cost requires finding the critical value using a t-distribution with 17 - 1 = 16 degrees of freedom. A t-score of approximately 1.746 is found for a 90% confidence level.

Step-by-step explanation:

To find the 90% confidence interval for the mean repair cost for the TVs, we first need to find the critical value. This involves using the t-distribution since the population standard deviation is unknown and the sample size is relatively small. We need a t-score that corresponds to a 90% confidence level for a two-tailed test with degrees of freedom (df) equal to the sample size minus one (n - 1).

To find the critical t-value, we use degrees of freedom df = 17 - 1 = 16 and look up the critical t-value for a 90% confidence interval in a t-distribution table or use a statistical software. However, if we assume that the population is approximately normal and the sample size is moderate (as in this case with 17 samples), the t-distribution becomes similar to the normal distribution, therefore sometimes a z-score can be used as an approximation. Nevertheless, it is advisable to use a t-score for more accuracy.

If rounding to three decimal places, the t-score for df = 16 at 90% confidence is approximately 1.746 (using a t-distribution table or software).

User Dzinx
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5.1k points
3 votes

Answer:

49.34 i think

Step-by-step explanation:

User Rahul Uttarkar
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4.9k points