Question

Warren County Telephone Company claims in its annual report that “the typical customer spends $61 per month on local and long-distance service.” A sample of 12 subscribers revealed the following amounts spent last month.

$63

$64

$64

$68

$60

$64

$69

$62

$64

$57

$54

$67

a.	What is the point estimate of the population mean? (Round your answer to 3 decimal places.)

Estimated population mean

$

b.	Develop a 90% confidence interval for the population mean. (Use t Distribution Table.) (Round your answers to 3 decimal places.)

Confidence interval for the population mean

$ and $

c.	Is the company's claim that the "typical customer" spends $61 per month reasonable?
	Yes No

Answer 1

Solution :

Given that 63,64,64,68,60,64,69,62,64,57,54,67

=> mean x-bar = sum of terms/number of terms

= 756/12

= 63

=> The point estimate of the population mean is 63.000

b. => n = 12 , mean x-bar = 63

=> standard deviation s = 4.349

=> df = n - 1 = 12 - 1 = 11

=> for 90% confidence interval , t = 1.796

=> The 90% confidence interval for the population mean is

=> x-abr +/- t*s/sqrt(n)

=> 63 +/- 1.796*4.349/sqrt(12)

=> (60.745 , 65.255)

=> Confidence interval for the population mean is $60.745 and $65.255

c. => answer:- "yes" , $61 is reasonable because it is lies in the confidence interval