Question

The table below shows the lengths of some randomly chosen CDs in John’s very large collection. Give the 92% confidence interval for the population mean, assuming that the population is approximately normally distributed. Round the endpoints to two decimal places.

69.33	74.64	65.28	61.01	38.52	69.07	40.01	42.14	41.52	56.76
58.14	70.80	39.21	71.36	46.94	52.90	56.16	41.01	56.57	60.97
64.62	69.12	66.58	58.26	62.79

Confidence interval:

Answer 1

x	x2
69.33	4806.6489
74.64	5571.1296
65.28	4261.4784
61.01	3722.2201
38.52	1483.7904
69.07	4770.6649
40.01	1600.8001
42.14	1775.7796
41.52	1723.9104
56.76	3221.6976
58.14	3380.2596
70.8	5012.64
39.21	1537.4241
71.36	5092.2496
46.94	2203.3636
52.9	2798.41
56.16	3153.9456
41.01	1681.8201
56.57	3200.1649
60.97	3717.3409
64.62	4175.7444
69.12	4777.5744
66.58	4432.8964
58.26	3394.2276
62.79	3942.5841
∑x=1433.71	∑x2=85438.7653

Mean ˉx=∑xn

=69.33+74.64+65.28+61.01+38.52+69.07+40.01+42.14+41.52+56.76+58.14+70.8+39.21+71.36+46.94+52.9+56.16+41.01+56.57+60.97+64.62+69.12+66.58+58.26+62.79/25

=1433.71/25

=57.3484

Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√85438.7653-(1433.71)22524

=√85438.7653-82220.974624

=√3217.790724

=√134.0746

=11.5791

Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 92% confidence level the t is ,

= 1 - 92% = 1 - 0.92= 0.08

/ 2 = 0.08 / 2 = 0.04

t /2,df = t0.04,24 =1.828

Margin of error = E = t/2,df * (s /n)

= 1.828* (11.58 / 25)

= 4.23

Margin of error = 4.23

The 92% confidence interval estimate of the population mean is,

- E < < + E

57.35 - 4.23 < < 57.35 + 4.23

53.12 < < 61.58

(53.12, 61.58)