Starting salaries of 64 college graduates who have taken a statistics course have a mean of $44,500 with a standard deviation of $6,800. find a 68% confidence interval for μ. (note: do not use commas or dollar signs in your answers. round each bound to three decimal places.)

Using the t-distribution, it is found that the 68% confidence interval for μ is  (43647.96, 45352.04).

We are given the standard deviation for the sample, which is why the t-distribution is used to solve this question.

The information given is:

The confidence interval is:

The critical value, using a t-distribution calculator, for a two-tailed 68% confidence interval, with 64 - 1 = 63 df, is t = 1.0024.

Then, the interval is:

The 68% confidence interval for μ is (43647.96, 45352.04).

A similar problem is given at link

The 68% confidence interval for μ is (43648.300, 45351.700)

Step-by-step explanation:

We have been given the following data:

Sample size = n = 64

Sample mean = = 44,500

Sample Standard Deviation = s = 6,800

We have to find 68% confidence interval for the given data. Since, value of Population Standard Deviation is unknown, and we have the value of Sample Standard Deviation, we will use t-distribution to find the required confidence interval.

The formula to calculate the confidence interval is:

Here,

is the critical t-value for 68% confidence level and n - 1 = 63 degrees of freedom. From the t-table, this critical value comes out to be t = 1.002

Using the values, we get the confidence interval:

Thus, the 68% confidence interval for μ is (43648.300, 45351.700)

true? is that what you're asking or do you mean something else?

the answer is c just got it on ed

step-by-step explanation: