In: Statistics and Probability
| 
 Congo Incorporated claimed in its annual report that "the typical customer spends $65 per month on streaming apps." A sample of 12 subscribers revealed the following amounts spent last month.  | 
| 
 $57  | 
 $69  | 
 $60  | 
 $68  | 
 $66  | 
 $65  | 
 $63  | 
 $64  | 
 $67  | 
 $59  | 
 $58  | 
 $58  | 
| (a) | 
 What is the point estimate of the population mean? (Round your answer to 3 decimal places.)  | 
| Population mean | $ | 
| (b) | 
 Determine the 90% confidence interval for the population mean. (Round your answers to 2 decimal places.)  | 
| Confidence interval | $ and $ | 
| (c) | Is the company's claim that the "typical customer" spends $65 per month reasonable? | ||||
  | 
Sample size : Number of subscribers in the sample : n= 12
Sample data : xi : Amount spent by the ith subscriber
(a) point estimate of the population mean is the sample mean :
Sample mean :

point estimate of the population mean = 62.833
(b)
Formula for Confidence Interval for Population mean when population Standard deviation is not known

s: Sample standard deviation

| xi | 
![]()  | 
![]()  | 
| 57 | -5.8333 | 34.0278 | 
| 69 | 6.1667 | 38.0278 | 
| 60 | -2.8333 | 8.0278 | 
| 68 | 5.1667 | 26.6944 | 
| 66 | 3.1667 | 10.0278 | 
| 65 | 2.1667 | 4.6944 | 
| 63 | 0.1667 | 0.0278 | 
| 64 | 1.1667 | 1.3611 | 
| 67 | 4.1667 | 17.3611 | 
| 59 | -3.8333 | 14.6944 | 
| 58 | -4.8333 | 23.3611 | 
| 58 | -4.8333 | 23.3611 | 
| 201.6667 | 

for 90% confidence level = (100-90)/100 =0.1
/2
= 0.05
Degrees of freedom = n-1 =12-1 =11
t
/2,n-1
= t0.05,11 = 1.7959
90% confidence interval for the population mean :

90% confidence interval for the population mean = (60.6132 , 65.0528)
(c) Is the company's claim that the "typical customer" spends $65 per month reasonable?
As 65 is within (60.6132 < 65 < 65.0528) the confidence interval ; Yes, the company's claim that the "typical customer" spends $65 per month reasonable
Answer : Yes