Final answer:
To find the 95% confidence interval for the mean repair cost for TVs, the critical value from the t-distribution is approximately 2.080, used with a standard error calculated as $4.84. The 95% confidence interval is then constructed and is ($73.16, $93.30).
Step-by-step explanation:
The task is to determine the 95% confidence interval for the mean repair cost for TVs, given that a consumer affairs investigator records a sample mean of $83.23 and standard deviation of $22.67 based on 22 randomly selected TVs.
Step 1: Find the critical value for the confidence interval.
Since the sample size is relatively small (n=22), and assuming the population distribution is approximately normal, we will use the t-distribution to find our critical value. The degrees of freedom (df) are n - 1, which for our sample is 21. For a 95% confidence interval, the two-tailed critical t-value for df=21 is approximately 2.080. This value can be found using statistical software or a t-distribution table.
Step 2: Construct the 95% confidence interval.
To construct the confidence interval, we use the formula for a t-confidence interval for the mean:
Confidence Interval = Sample mean ± (Critical value × Standard error)
The standard error (SE) is the standard deviation divided by the square root of the sample size, which gives:
SE = $22.67 / √22 ≈ $4.84
Now, we can calculate the margin of error (ME):
ME = Critical value × SE ≈ 2.080 × $4.84 ≈ $10.07
With these values, the 95% confidence interval is:
($83.23 - $10.07, $83.23 + $10.07) = ($73.16, $93.30)
Therefore, we are 95% confident that the true population mean for the repair cost of TVs lies between $73.16 and $93.30.