In: Statistics and Probability
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:
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)