Question
In: Statistics and Probability
The table below shows the lengths of some randomly chosen CDs in John’s very large collection....
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:
Solutions
Expert Solution
| 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 /
= 1.828* (11.58 /
= 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)
orchestra answered 1 year agoRelated Solutions
The table below shows the lengths of some randomly chosen CDs in John’s very large collection....
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: ( , )
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