Question

DO PODE (10 ) Exercise 3 Consider the data below that correspond to the monthly talk times in minutes) of the mobile phones of 12 employees of one department (Customer Service) of a certain company (assume that talk times are normally distributed): 123 88 42 91 77 66 97 117 110 85 53 38 a) Estimate the mean monthly talk time of all employees in all departments of the company i.e. calculate the best one-point estimate for the population mean). (10 marks) b) Estimate a 95% confidence interval for the mean monthly talk time of all employees of the company. (10 marks) c) Do you think that this sample of data is representative? Why or why not? (10 marks)

Answer 1

Solution-A:

best point of population mean =sample mean

sample mean=sum of monthly talk time/total values

=123+42+77+97+110+53+88+91+66+117+85+38/

= 987/12

=82.25

Solution-B:

X	mean	X-MEAN	(X-MEAN)^2
123	82.25	40.75	1660.563
42	82.25	-40.25	1620.063
77	82.25	-5.25	27.5625
97	82.25	14.75	217.5625
110	82.25	27.75	770.0625
53	82.25	-29.25	855.5625
88	82.25	5.75	33.0625
91	82.25	8.75	76.5625
66	82.25	-16.25	264.0625
117	82.25	34.75	1207.563
85	82.25	2.75	7.5625
38	82.25	-44.25	1958.063
		TOTAL	8698.25

Standard deviation

s= $\sqrt{(x-xbar)^2/n-1}$

s=sqrt(8698.25/12-1)

s=28.12028

95% confidence interval for mean monthly talk time of all employees is

xbar-t*s/sqrt(n),xbar+t*s/sqrt(n)

t crit for 11 df and 0.025 is

=T.INV(0.025,11)=2.20099

so 95% confidence interval for mean monthly talk time is

82.25-2.20099*28.12028/sqrt(12),82.25+2.20099*28.12028/sqrt(12),

64.38319, 100.1168

95% lower limit for mean monthly talk time=64.38319

95% upper limit for mean monthly talk time=100.1168

Solution-c:

yes its representative as its random sample