Question

Salaries of 45 college graduates who took a statistics course in college have a​ mean, x overbar​, of $ 61 comma 300. assuming a standard​ deviation, sigma​, of ​$18 comma 246​, construct a 99​% confidence interval for estimating the population mean mu.

Answer 1

We are given:
sample size= n = 45
sample mean = x = 61,300
sample standard deviation =18,246

Since the population standard deviation is not known we will use t distribution to find the confidence interval.

Confidence interval = 99%
Degrees of freedom= n - 1 = 45 – 1 = 44
Critical t value = 2.692

The 99% confidence interval will be:$(x-t_{critical}* \frac{s}{ \sqrt{n} }, x+t_{critical}* \frac{s}{ \sqrt{n} }) \\ \\ (61300-2.692* \frac{18246}{ \sqrt{45} }, 61300+2.692* \frac{18246}{ \sqrt{45} }) \\ \\ (53978,68622)$

Thus, the 99% confidence interval for the population mean is:
53978 to 68622

Answer 2

Answer:

(53978, 68622)

Step-by-step explanation:

Mean = 61,300

Std deviation = 18,246

Sample size  n = 45

Std error of sample = s/\sqrt n

=2719.95

Only sample std devitaion is known so we can use t critical value for df 44

t critical for 99% for df 44 = 2.692

Margin of error = 2.692(SE)

=7322

Confidence interval lower bound = 61300-7322 = 53978

Upper limit = 61300+7322 = 68622