Question

A student records the repair cost for 17 randomly selected washers. A sample mean of $82.95 and standard deviation of $14.89 are subsequently computed. Determine the 80% confidence interval for the mean repair cost for the washers. 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._____Step 2 of 2: Construct the 80% confidence interval. Round your answer to two decimal places.Lower Endpoint:______ Upper Endpoint:_______

Answer 1

To construct an 80% confidence interval, first find the critical t-value, which is approximately 1.337 for 16 degrees of freedom. Then, calculate the lower and upper interval endpoints, which are approximately $79.08 and $86.82, respectively.

Step-by-step explanation:

To construct an 80% confidence interval for the mean repair cost for washers based on the given sample data, we follow two steps:

1. Find the critical value: For an 80% confidence level, we consult the t-distribution table looking for a critical value that leaves 10% in the upper tail (since it's a two-tailed test) for 16 degrees of freedom (n - 1, where n is the sample size of 17). The critical value (t) to three decimal places is approximately 1.337.
2. Construct the confidence interval: The formula for a confidence interval is mean ± (critical value) × (standard deviation / √(n)). This yields the lower and upper endpoints as follows:

So, the 80% confidence interval for the mean repair cost is approximately ($79.08, $86.82).

Answer 2

Critical value: T = 1.34

Lower endpoint: $63.04

Upper endpoint: $102.86

Step-by-step explanation:

We are in posession of the sample's standard deviation, so we use the student t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 17 - 1 = 16

80% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 16 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.8)/(2) = 0.9([tex]t_(9)`). So we have T = 1.34 as our critical value.

The margin of error is:

M = T*s = 1.337*14.89 = 19.91.

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 82.95 - 19.91 = $63.04

The upper end of the interval is the sample mean added to M. So it is 82.95 + 19.91 = $102.86