Question

4. By using the information provided below, find a t–confidence interval for the mean of the population from which the sample was drawn. [4 points]
(a) ¯ x = 20, n = 36, s = 3, confidence level = 95%.

(b) ¯ x = 35, n = 25, s = 4, confidence level = 90%.

5. A 95% confidence interval for a population mean, µ, is given as (18.985,21.015). This confidence interval is based on a simple random sample of 36 observations. Calculate the sample mean and standard deviation. Assume that all conditions necessary for inference are satisfied. Use the t–distribution in any calculations. [4 points]

6. A market researcher wants to evaluate car insurance savings at a competing company. Based on past studies he is assuming that the standard deviation of savings is $100. He wants to collect data such that he can get a margin of error of no more than $10 at a 95% confidence level. How large of a sample should he collect

Answer 1

Solution:-

4. (a) x = 20, n = 36, s = 3, confidence level = 95%.

df = n-1 = 36-1 = 35 , t = 2.03

95% confidence interval for the mean = X +/- t*s/sqrt(n)

= 20 +/- 2.03*3/sqrt(36)

=  18.985,21.015

(b) x = 35, n = 25, s = 4, confidence level = 90%.

df = n-1 = 24, t = 1.7109

90% confidence interval for the mean = X +/- t*s/sqrt(n)

= 35 +/- 1.7109*4/sqrt(25)

= 33.6313 , 36.3687

5.

sample mean = (18.985+21.015)/2 =20

margin of error = 20 -18.985 = 1.015

The degree of freedom =n-1=36-1=35

Given a=0.05, t(0.025, df=35) =2.03 (from student t table)

Since margin of error = t*s/vn =1.0285

---> 2.03*s/6 =1.0285

So standard deviation = 1.0285*6/2.03 =3.039901

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6. n = (Z*s/E)^2 = (1.96*100/10)^2 = 384.16 = 384