Final answer:
To calculate the 95% confidence interval for the mean repair cost of computers given a sample mean of $100, standard deviation of $42.50, and a sample size of 15, we use a t-distribution due to the small sample size and unknown population standard deviation. After finding the correct t-score for 14 degrees of freedom, we calculate the margin of error and apply it to the sample mean to find that the interval is approximately between $76.46 and $123.54.
Step-by-step explanation:
To construct a 95% confidence interval for the mean repair cost of computers when the sample mean is $100, and the standard deviation is $42.50 with a sample size of 15, we must first recognize that we will be using the t-distribution since we are using the sample standard deviation instead of a known population standard deviation and our sample size is relatively small.
We begin by using the formula for the confidence interval:
CI = μ ± (t* × (s/√n))
Where μ is the sample mean, t* is the t-score corresponding to a 95% confidence degree of freedom (n-1), s is the sample standard deviation, and n is the sample size.
To find the appropriate t-score, we look up the value in a t-distribution table or use a software/calculator with a degrees of freedom of (15-1) or 14. After finding the t-score, we calculate the margin of error (t* × (s/√n)).
Finally, we add and subtract this margin of error from our sample mean to get the lower and upper bounds of our confidence interval. For illustrative purposes, let us assume the t-score is approximately 2.145 (this should be verified with a t-table or software for accurate calculations).
Margin of Error = 2.145 × ($42.50/√15)
= 2.145 × 10.98
= $23.54
Our 95% confidence interval is therefore:
CI = $100 ± $23.54
= ($76.46, $123.54)
This means we are 95% confident that the average repair cost of computers lies between $76.46 and $123.54.