Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

1.A random sample of 70 observations produced a mean of x=31.5x from a populatio

ID: 3134804 • Letter: 1

Question

1.A random sample of 70 observations produced a mean of x=31.5x from a population with a normal distribution and a standard deviation =2.45

(a) Find a 95% confidence interval for

(b) Find a 99% confidence interval for

c) Find a 90% confidence interval for

2. Starting salaries of 130 college graduates who have taken a statistics course have a mean of $44,609. Suppose the distribution of this population is approximately normal and has a standard deviation of $10,228.

Using an 85% confidence level, find both of the following:

(a) The margin of error:

(b) The confidence interval for the mean :

3. A random sample of nn measurements was selected from a population with standard deviation =15 and unknown mean . Calculate a 90% confidence interval for for each of the following situations:

(a) n=55, x=84.8

(b)  n=85, x=84.8

(c)  n=115, x=84.8

(d) In general, we can say that for the same confidence level, increasing the sample size ____________ the margin of error (width) of the confidence interval

4. What is the value of t*, the critical value of the t distribution for a sample of size 18, such that the probability of being greater than t* is 1%?

5.What is the value of t*, the critical value of the t distribution for a sample of size 14, such that the probability of being greater than t* or less than -t* is 1%?

Please and thank you!!!

Explanation / Answer

Answer to question# 1)

Formula of confidence interval is :

x bar - z* SE , xbar + z*SE

.

We got x bar = 31.5

SE = 2.45 / sqrt(70)

SE = 0.2928

.

Answer to part a)

For 95% confidence interval , z = 1.96

.

On plugging the values weget:

31.5 - 1.96 * 0.2928 , 31.5 + 1.96 *0.28928

30.9261 , 32.067

.

Answer to part b)

Z value for 99% confidence level is: 2.58

31.5 - 2.58 *0.2928 , 31.5 + 2.58 *0.2928

30.7446 , 32.2554

.

Answer to part c)

Z value for 90% confidence interval is 1.645

31.5 -1.645 *0.2928 , 31.5 +1.645 *0.2928

31.0183 , 31.9817