4. A population consists of the following five values: 10, 11, 12, 17 and 20;
consider sampling without replacement from this population.
• List all samples of size three, and compute the mean of each sample.
• Compute the mean of the distribution of sample means, and the
population mean, and compare these two values.
• Compare and comment on the spread in the population with that in
the sample means.
5. If you select a sample of size n = 40 from a Normal population with a
mean of 75 and a standard deviation of 5, find the probabilities that your
sample mean is
• Less than 74
• Between 74 and 76
• Between 76 and 77
• More than 77
Recalculate each of these probabilities if you double your sample size.
6. Assuming that the following sample of 12 numbers arise from a Normal
distribution with mean μ and standard deviation σ = 5.2, obtain 95%, 99%
and 99.9% confidence intervals for μ.
91.3 92.4 101.1 99.5 104.7 95.5
97.2 106.3 93.7 108.1 97.9 90.2
7. For the data in Q6, evaluate the probability or p-value for testing the null
hypothesis H0 : μ = 100 against the two-sided alternative hypothesis
H1 : μ = 100. What evidence do the CIs in Q6 and this p-value offer in
support of H0?
Normal distribution 20點 急急急 !!
2010-03-15 12:28 pm
回答 (1)
2010-03-17 10:36 pm
✔ 最佳答案
4a:(10, 11, 12), (10, 12, 11), (11, 10, 12), (11, 12, 10), (12, 10, 11), (12, 11, 10) ==> mean = 11
(10, 11, 17), (10, 17, 11), (11, 10, 17), (11, 17, 10), (17, 10, 11), (17, 11, 10) ==> mean = 12.7
(10, 11, 20), (10, 20, 11), (11, 10, 20), (11, 20, 10), (20, 10, 11), (20, 11, 10) ==> mean = 13.7
(10, 12, 17), (10, 17, 12), (12, 10, 17), (12, 17, 10), (17, 10, 12), (17, 12, 10) ==> mean = 13
(10, 12, 20), (10, 20, 12), (12, 10, 20), (12, 20, 10), (20, 10, 12), (20, 12, 10) ==> mean = 14
(11, 12, 17), (11, 17, 12), (12, 11, 17), (12, 17, 11), (17, 11, 12), (17, 12, 11) ==> mean = 13.3
(11, 12, 20), (11, 20, 12), (12, 11, 20), (12, 20, 11), (20, 11, 12), (20, 12, 11) ==> mean = 14.3
(10, 17, 20), (10, 20, 17), (17, 10, 20), (17, 20, 10), (20, 10, 17), (20, 17, 10) ==> mean = 15.7
(11, 17, 20), (11, 20, 17), (17, 11, 20), (17, 20, 11), (20, 11, 17), (20, 17, 11) ==> mean = 16
(12, 17, 20), (12, 20, 17), (17, 12, 20), (17, 20, 12), (20, 12, 17), (20, 17, 12) ==> mean = 16.3
4b: The overall sample mean = (11 + 12.7 + 13 + 13.3 + 13.7 + 14 + 14.3 + 15.7 + 16 + 16.3) / 10 = 14
The population mean = (10 + 11 + 12 + 17 + 20) / 5 = 14
They are the same.
4c: population variance = 14.8
variance of sample mean = 2.467 (actually, you could theoretically prove that variance of sample mean = population variance / sample size, when the size of population is large)
The variance of sample mean is much smaller than the population variance.
2010-03-17 14:36:32 補充：
5: sample mean = 75, standard deviation of sample mean = 5 / sqrt(40) = 0.79.
If we double the sample size, sample mean = 75, standard deviation of sample mean = 5 / sqrt(80) = 0.56
2010-03-17 14:36:41 補充：
5ai: sample size = 40, Pr (mean < 74) = Pr [Z < (74 – 75) / 0.79] = Pr (Z < -1.2649) = 0.10295
5aii: sample size = 80, Pr (mean < 74) = Pr [Z < (74 – 75) / 0.56] = Pr (Z < -1.7889) = 0.03682
2010-03-17 14:36:47 補充：
5bi: sample size = 40, Pr (74< mean < 76) = Pr [(74 – 75) / 0.79 < Z < (76 – 75) / 0.79] = Pr (-1.2649< Z < 1.2649) = 0.79410
5bii: sample size = 80, Pr (74< mean < 76) = Pr [(74 – 75) / 0.56 < Z < (76 – 75) / 0.56] = Pr (-1.7889 < Z < 1.7889) = 0.92636
2010-03-17 14:36:58 補充：
5ci: sample size = 40, Pr (76< mean < 77) = Pr [(76 – 75) / 0.79 < Z < (77 – 75) / 0.79] = Pr (1.2649< Z < 2.5298) = 0.09725
5cii: sample size = 80, Pr (76< mean < 77) = Pr [(76 – 75) / 0.56 < Z < (77 – 75) / 0.56] = Pr (1.7889 < Z < 3.5777) = 0.03665
2010-03-17 14:37:09 補充：
5di: sample size = 40, Pr (mean > 77) = Pr [Z < (77 – 75) / 0.79] = Pr (Z < 2.5298) = 0.00571
5dii: sample size = 80, Pr (mean > 77) = Pr [Z < (77 – 75) / 0.56] = Pr (Z < 3.5777) = 0.00017
2010-03-17 14:37:24 補充：
6: sample mean = 98.158
6a: 95%
98.158 – 1.96 x 5.2 / sqrt(12) < population mean < 98.158 + 1.96 x 5.2 / sqrt(12)
95.2162 < population mean < 101.1005
2010-03-17 14:37:31 補充：
6b: 99%
98.158 – 2.5758 x 5.2 / sqrt(12) < population mean < 98.158 + 2.5758 x 5.2 / sqrt(12)
94.2917 < population mean < 102.0249
2010-03-17 14:37:40 補充：
6c: 99.9%
98.158 – 3.2905 x 5.2 / sqrt(12) < population mean < 98.158 + 3.2905 x 5.2 / sqrt(12)
93.2189 < population mean < 103.0978
2010-03-17 14:37:47 補充：
7: H0: mean = 100
Pr [Z < sqrt(12) x (100 – 98.158) / 5.2] = 78% confidence
2010-03-17 14:38:24 補充：
i.e. around 22% probability that the population is not 100
收錄日期: 2021-04-23 18:26:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100315000051KK00261