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Sagot :
Answer:
HI BRAINLY USER!
A study of 36 camels showed that they could walk an average rate of 4.2 kilometers per hour. The sample standard deviation is 0.64. Find the 90% confidence interval of the mean for all camels
90% CI for using Normal dist
Sample Mean=7=4.2
Since sample size is large, sample standard deviation can be taken as an approx-
imation of population standard deviation.
Population Standard deviation =⁰ = 0.64
Sample Size = n = 36
Significance level = a = 1- Confidence 1-0.9 = 0.1
The Critical Value = a/2 = 20.05 = 1.645 (From z table, using interpolation,1/2 th distance between 1.64 and 1.65 )
Critical Values, a/2 = ±1.645
Margin of Error, E= 0.175)
Limits of 90% confidence interval are given by:
90% confidence interval is: E=4.2±0.175467
= (4.024533, 4.375467)
90% CI using normal dist: 4.025 < p < 4.375
Also, the exact answer using TI-83/84 or excel is
4.2+ invNorm(0.95) x 0.64/sqrt(36)
= (4.0245483982.4.3754516018)
Note that some professor prefer 1-distribution when standard deviation is un known even though sample size is more than 30, in this case the answer will be: 90% CI using t-dist: 4.020 < p < 4.380
3. If the standard deviation of the lifetimes of television tubes is estimated as 100 hours, how large a sample must be selected in order to be 90% confident that the error in the estimated mean lifetime will not exceed 20 hours?
exceed 20 hours.
at least 68
10 hours
ANS: (b) at least 271
Step-by-step explanation:
QUESTION
A study of 36 camels showed that they could walk an average rate of 4.2 kilometers per hour. The sample standard deviation is 0.64. Find the 90% confidence interval of the mean for all camels
___________________________________
[tex]ANSWER & SOLUTION[/tex]
90% CI for using Normal dist
Sample Mean=7=4.2
Since sample size is large, sample standard deviation can be taken as an approx
imation of population standard deviation.
Population Standard deviation = ⁰ = 0.64
- Sample Size = n = 36
- Significance level = a = 1- Confidence 1 0.9 = 0.1
- The Critical Value = a/2 = 20.05 = 1.645
(From z table, using interpolation,1/2 th distance
between 1.64 and 1.65 )
Critical Values, a/2 = ±1.645
[tex]Marginof Error E = za/2 \times = \sqrt[0]{n} = 1.645 = \sqrt[0.64]{36} ≈0.175467[/tex]
Margin of Error, E= 0.175)
Limits of 90% confidence interval are given by:
[tex]Lowerlimit -E-4.2 - 0.1754674.0245334.025[/tex]
[tex]Upperlimit=7+ E= 4.2+ 0.1754674.3754674.375[/tex]
- 90% confidence interval is: E=4.2+0.175467
- = (4.024533, 4.375467)
90% CI using normal dist: 4.025 < p < 4.375
Also, the exact answer using TI-83/84 or excel is
- ➝ 4.2+ invNorm(0.95) x 0.64/sqrt(36)
- = (4.0245483982.4.3754516018)
__________________________________
Note : that some professor prefer 1 distribution when standard deviation is un known even though sample size is more than 30, in this case the answer will be: 90% CI using t-dist: 4.020 < p < 4.380
#CARRY ON LEARNING
– zhen
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