SOLUTION: Jim’s Auto Insurance Company customers sometimes have to wait a long time to speak to a customer service representative when they call regarding disputed claims. A random sample

Question 1201334: Jim’s Auto Insurance Company customers sometimes have to wait a long time to speak to a customer service representative when they call regarding disputed claims. A random sample of 25 such calls yielded a mean waiting time of 22 minutes with a standard deviation of 6 minutes. Construct a 99% confidence Interval for the population mean of such waiting times. Assume that such waiting times for the population follow a normal distribution.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

mu = population mean
sigma = population standard deviation

n = 25 = sample size
xbar = 22 = sample mean
s = 6 = sample standard deviation

Because we don't know the value of sigma, and because n > 30 isn't the case, we must use the T distribution.

df = degrees of freedom
df = n-1
df = 25-1
df = 24

At 99% confidence and df = 24, the t critical value is roughly t = 2.797
Use a table like this
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
to get that value.
Look at the row labeled df = 24.
Also, look at the column labeled "confidence level 99%".
The confidence levels are labeled at the bottom.

Using a T distribution calculator is also another option to determine the t critical value.

In summary so far, we have these input values:

xbar = 22
s = 6
n = 25
t = 2.797 approximately

Let's compute the margin of error for the mean.
E = margin of error
E = t*s/sqrt(n)
E = 2.797*6/sqrt(25)
E = 3.3564
This value is approximate.

Now we can compute the boundaries.
L = lower boundary of the confidence interval
L = xbar - E
L = 22 - 3.3564
L = 18.6436
and
U = upper boundary of the confidence interval
U = xbar + E
U = 22 + 3.3564
U = 25.3564
These values are approximate.

The 99% confidence interval in the format (L, U) is approximately (18.6436, 25.3564)

The 99% confidence interval in the format L < mu < U is approximately 18.6436 < mu < 25.3564
This second format is a bit more descriptive in terms of which population parameter we're trying to measure.

Therefore, we are 99% confident the population mean waiting time is somewhere between roughly 18.6436 minutes and 25.3564 minutes.

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