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Shuangshuang · 2021年02月28日

第二问为什么是负的2.32呢

NO.PZ2020010304000053

问题如下:

A data management group wants to test the null hypothesis that observed data is N(0,1) distributed by evaluating the mean of a set of random draws. However, the actual underlying data is distributed as N(1, 2.25).

a. If the sample size is 10, what is the probability of a Type II error and the power of the test? Assume a 90% confidence level on a two-sided test.

b. How many samples would need to be taken to reduce the probability of a Type II error to less than 1%?

解释:

a. When the null hypothesis is false, the probability of a Type II error is equal to the probability that the hypothesis fails to be rejected.

Now, if there are 10 samples taken from an N(0,1) then the standard deviation is reduced

σH0=1/10=0.316\sigma_{H_0}=1/ \sqrt{10}=0.316

Therefore, the cut-off points are ±1.650.316=±0.522\pm1.65 * 0.316 = \pm0.522

In actuality, the true distribution is N(1,2.25), so the σ=22.25=1.5\sigma = \sqrt{22.25} = 1.5. For a sample size of 10, the expected sample standard deviation is

σsample=1.5/10=0.474\sigma_{sample}=1.5/ \sqrt{10} = 0.474

Calculating the equivalent distance of ±1.65\pm1.65 in this distribution compared to a standard N(0,1) yields

left = (-0.522-1)/0.474=-3.21

and right =(+0.522 - 1)/ 0.474 = -1.00

The probability of being on the left-hand side is practically zero. For the right, Pr(> right) = 1 - Φ(-1.00) = 1 - 15.9% = 84.1%.

So total probability of a Type II error is 1 – the probability of being in the two tails is

Pr(Non - Rejection|Ho is false) = 1 - [Pr(< left) + Pr(> right)] ≈ 1 - 84.1% = 15.9%

Therefore, the power of the test is 84.1%.

b. The requirement is to have 1 - [Pr(< left) + Pr(> right)] = 1%

Clearly, as n increases from 10, the probability of being in the left-hand tail will only decrease from already being close to zero.

Therefore, the requirement becomes 1 - Pr(> right) = 0.01

This occurs at a Z-score of (using the Excel function NORMSINV) -2.32.

Accordingly, the following equations need to be solved

1.65σH0=1.65/n=K1.65 * \sigma_{H_0} =1.65/ \sqrt n=K and

+K11.5n=2.32\frac{+K-1}{\frac{1.5}{\sqrt n}}=-2.32

Plugging in K yields:

1.65n11.5n=1.65n1.5=2.32n=5.13\frac{\frac{1.65}{\sqrt n}-1}{\frac{1.5}{\sqrt n}}= \frac{1.65-\sqrt n}{1.5}=-2.32\geq \sqrt n=5.13 n=26.3\geq n=26.3

And because partial observations are not allowed, n = 27.

为神马第二问标准化后是-2.33呢
1 个答案

品职答疑小助手雍 · 2021年03月01日

嗨,从没放弃的小努力你好:


因为问的是第二类错误发生小于1%的概率,也就意味着Pr(> right) = 0.99,所以这个right对应的应该是-2.33(根据特殊的分位数,这个数应该要记得)

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NO.PZ2020010304000053 问题如下 A ta management group wants to test the null hypothesis thobserveta is N(0,1) stributeevaluating the meof a set of ranm aws. However, the actuunrlying ta is stributeN(1, 2.25). If the sample size is 10, whis the probability of a Type II error anthe power of the test? Assume a 90% confinlevel on a two-sitest.How many samples woulneeto taken to rethe probability of a Type II error to less th1%? When the null hypothesis is false, the probability of a Type II error is equto the probability ththe hypothesis fails to rejecteNow, if there are 10 samples taken from N(0,1) then the stanrviation is receH0=1/10=0.316\sigma_{H_0}=1/ \sqrt{10}=0.316σH0​​=1/10​=0.316Therefore, the cut-off points are ±1.65∗0.316=±0.522\pm1.65 * 0.316 = \pm0.522±1.65∗0.316=±0.522In actuality, the true stribution is N(1,2.25), so the σ=22.25=1.5\sigma = \sqrt{22.25} = 1.5σ=22.25​=1.5. For a sample size of 10, the expectesample stanrviation isσsample=1.5/10=0.474\sigma_{sample}=1.5/ \sqrt{10} = 0.474σsample​=1.5/10​=0.474Calculating the equivalent stanof ±1.65\pm1.65±1.65 in this stribution compareto a stanrN(0,1) yielleft = (-0.522-1)/0.474=-3.21anright =(+0.522 - 1)/ 0.474 = -1.00The probability of being on the left-hansi is practically zero. For the right, Pr( right) = 1 - Φ(-1.00) = 1 - 15.9% = 84.1%.So totprobability of a Type II error is 1 – the probability of being in the two tails isPr(Non - Rejection|Ho is false) = 1 - [Pr( left) + Pr( right)] ≈ 1 - 84.1% = 15.9%Therefore, the power of the test is 84.1%.The requirement is to have 1 - [Pr( left) + Pr( right)] = 1%Clearly, n increases from 10, the probability of being in the left-hantail will only crease from alrea being close to zero.Therefore, the requirement becomes 1 - Pr( right) = 0.01This occurs a Z-score of (using the Excel function NORMSINV) -2.32.Accorngly, the following equations neeto solve.65∗σH0=1.65/n=K1.65 * \sigma_{H_0} =1.65/ \sqrt n=K1.65∗σH0​​=1.65/n​=K an+K−11.5n=−2.32\frac{+K-1}{\frac{1.5}{\sqrt n}}=-2.32n​1.5​+K−1​=−2.32Plugging in K yiel: 1.65n−11.5n=1.65−n1.5=−2.32≥n=5.13\frac{\frac{1.65}{\sqrt n}-1}{\frac{1.5}{\sqrt n}}= \frac{1.65-\sqrt n}{1.5}=-2.32\geq \sqrt n=5.13n​1.5​n​1.65​−1​=1.51.65−n​​=−2.32≥n​=5.13 ≥n=26.3\geq n=26.3≥n=26.3Anbecause partiobservations are not allowe n = 27. 为什么totprobability of a Type II error is 1 – the probability of being in the two tailsP(Type II error) = P(H0假 接受H0)/P(H0假)=1-P(H0假 拒绝H0)/P(H0假)但我觉得the probability of being in the two tails = P(H0假 拒绝H0)啊

2024-05-05 13:48 4 · 回答

NO.PZ2020010304000053 问题如下 A ta management group wants to test the null hypothesis thobserveta is N(0,1) stributeevaluating the meof a set of ranm aws. However, the actuunrlying ta is stributeN(1, 2.25). If the sample size is 10, whis the probability of a Type II error anthe power of the test? Assume a 90% confinlevel on a two-sitest.How many samples woulneeto taken to rethe probability of a Type II error to less th1%? When the null hypothesis is false, the probability of a Type II error is equto the probability ththe hypothesis fails to rejecteNow, if there are 10 samples taken from N(0,1) then the stanrviation is receH0=1/10=0.316\sigma_{H_0}=1/ \sqrt{10}=0.316σH0​​=1/10​=0.316Therefore, the cut-off points are ±1.65∗0.316=±0.522\pm1.65 * 0.316 = \pm0.522±1.65∗0.316=±0.522In actuality, the true stribution is N(1,2.25), so the σ=22.25=1.5\sigma = \sqrt{22.25} = 1.5σ=22.25​=1.5. For a sample size of 10, the expectesample stanrviation isσsample=1.5/10=0.474\sigma_{sample}=1.5/ \sqrt{10} = 0.474σsample​=1.5/10​=0.474Calculating the equivalent stanof ±1.65\pm1.65±1.65 in this stribution compareto a stanrN(0,1) yielleft = (-0.522-1)/0.474=-3.21anright =(+0.522 - 1)/ 0.474 = -1.00The probability of being on the left-hansi is practically zero. For the right, Pr( right) = 1 - Φ(-1.00) = 1 - 15.9% = 84.1%.So totprobability of a Type II error is 1 – the probability of being in the two tails isPr(Non - Rejection|Ho is false) = 1 - [Pr( left) + Pr( right)] ≈ 1 - 84.1% = 15.9%Therefore, the power of the test is 84.1%.The requirement is to have 1 - [Pr( left) + Pr( right)] = 1%Clearly, n increases from 10, the probability of being in the left-hantail will only crease from alrea being close to zero.Therefore, the requirement becomes 1 - Pr( right) = 0.01This occurs a Z-score of (using the Excel function NORMSINV) -2.32.Accorngly, the following equations neeto solve.65∗σH0=1.65/n=K1.65 * \sigma_{H_0} =1.65/ \sqrt n=K1.65∗σH0​​=1.65/n​=K an+K−11.5n=−2.32\frac{+K-1}{\frac{1.5}{\sqrt n}}=-2.32n​1.5​+K−1​=−2.32Plugging in K yiel: 1.65n−11.5n=1.65−n1.5=−2.32≥n=5.13\frac{\frac{1.65}{\sqrt n}-1}{\frac{1.5}{\sqrt n}}= \frac{1.65-\sqrt n}{1.5}=-2.32\geq \sqrt n=5.13n​1.5​n​1.65​−1​=1.51.65−n​​=−2.32≥n​=5.13 ≥n=26.3\geq n=26.3≥n=26.3Anbecause partiobservations are not allowe n = 27. 请问这个observeta是指什么?是抽样一次得到的数据么?

2024-04-28 13:41 1 · 回答

NO.PZ2020010304000053 问题如下 A ta management group wants to test the null hypothesis thobserveta is N(0,1) stributeevaluating the meof a set of ranm aws. However, the actuunrlying ta is stributeN(1, 2.25). If the sample size is 10, whis the probability of a Type II error anthe power of the test? Assume a 90% confinlevel on a two-sitest.How many samples woulneeto taken to rethe probability of a Type II error to less th1%? When the null hypothesis is false, the probability of a Type II error is equto the probability ththe hypothesis fails to rejecteNow, if there are 10 samples taken from N(0,1) then the stanrviation is receH0=1/10=0.316\sigma_{H_0}=1/ \sqrt{10}=0.316σH0​​=1/10​=0.316Therefore, the cut-off points are ±1.65∗0.316=±0.522\pm1.65 * 0.316 = \pm0.522±1.65∗0.316=±0.522In actuality, the true stribution is N(1,2.25), so the σ=22.25=1.5\sigma = \sqrt{22.25} = 1.5σ=22.25​=1.5. For a sample size of 10, the expectesample stanrviation isσsample=1.5/10=0.474\sigma_{sample}=1.5/ \sqrt{10} = 0.474σsample​=1.5/10​=0.474Calculating the equivalent stanof ±1.65\pm1.65±1.65 in this stribution compareto a stanrN(0,1) yielleft = (-0.522-1)/0.474=-3.21anright =(+0.522 - 1)/ 0.474 = -1.00The probability of being on the left-hansi is practically zero. For the right, Pr( right) = 1 - Φ(-1.00) = 1 - 15.9% = 84.1%.So totprobability of a Type II error is 1 – the probability of being in the two tails isPr(Non - Rejection|Ho is false) = 1 - [Pr( left) + Pr( right)] ≈ 1 - 84.1% = 15.9%Therefore, the power of the test is 84.1%.The requirement is to have 1 - [Pr( left) + Pr( right)] = 1%Clearly, n increases from 10, the probability of being in the left-hantail will only crease from alrea being close to zero.Therefore, the requirement becomes 1 - Pr( right) = 0.01This occurs a Z-score of (using the Excel function NORMSINV) -2.32.Accorngly, the following equations neeto solve.65∗σH0=1.65/n=K1.65 * \sigma_{H_0} =1.65/ \sqrt n=K1.65∗σH0​​=1.65/n​=K an+K−11.5n=−2.32\frac{+K-1}{\frac{1.5}{\sqrt n}}=-2.32n​1.5​+K−1​=−2.32Plugging in K yiel: 1.65n−11.5n=1.65−n1.5=−2.32≥n=5.13\frac{\frac{1.65}{\sqrt n}-1}{\frac{1.5}{\sqrt n}}= \frac{1.65-\sqrt n}{1.5}=-2.32\geq \sqrt n=5.13n​1.5​n​1.65​−1​=1.51.65−n​​=−2.32≥n​=5.13 ≥n=26.3\geq n=26.3≥n=26.3Anbecause partiobservations are not allowe n = 27. 这道题题干能不能翻译一下

2024-04-26 15:42 1 · 回答

NO.PZ2020010304000053 问题如下 A ta management group wants to test the null hypothesis thobserveta is N(0,1) stributeevaluating the meof a set of ranm aws. However, the actuunrlying ta is stributeN(1, 2.25). If the sample size is 10, whis the probability of a Type II error anthe power of the test? Assume a 90% confinlevel on a two-sitest.How many samples woulneeto taken to rethe probability of a Type II error to less th1%? When the null hypothesis is false, the probability of a Type II error is equto the probability ththe hypothesis fails to rejecteNow, if there are 10 samples taken from N(0,1) then the stanrviation is receH0=1/10=0.316\sigma_{H_0}=1/ \sqrt{10}=0.316σH0​​=1/10​=0.316Therefore, the cut-off points are ±1.65∗0.316=±0.522\pm1.65 * 0.316 = \pm0.522±1.65∗0.316=±0.522In actuality, the true stribution is N(1,2.25), so the σ=22.25=1.5\sigma = \sqrt{22.25} = 1.5σ=22.25​=1.5. For a sample size of 10, the expectesample stanrviation isσsample=1.5/10=0.474\sigma_{sample}=1.5/ \sqrt{10} = 0.474σsample​=1.5/10​=0.474Calculating the equivalent stanof ±1.65\pm1.65±1.65 in this stribution compareto a stanrN(0,1) yielleft = (-0.522-1)/0.474=-3.21anright =(+0.522 - 1)/ 0.474 = -1.00The probability of being on the left-hansi is practically zero. For the right, Pr( right) = 1 - Φ(-1.00) = 1 - 15.9% = 84.1%.So totprobability of a Type II error is 1 – the probability of being in the two tails isPr(Non - Rejection|Ho is false) = 1 - [Pr( left) + Pr( right)] ≈ 1 - 84.1% = 15.9%Therefore, the power of the test is 84.1%.The requirement is to have 1 - [Pr( left) + Pr( right)] = 1%Clearly, n increases from 10, the probability of being in the left-hantail will only crease from alrea being close to zero.Therefore, the requirement becomes 1 - Pr( right) = 0.01This occurs a Z-score of (using the Excel function NORMSINV) -2.32.Accorngly, the following equations neeto solve.65∗σH0=1.65/n=K1.65 * \sigma_{H_0} =1.65/ \sqrt n=K1.65∗σH0​​=1.65/n​=K an+K−11.5n=−2.32\frac{+K-1}{\frac{1.5}{\sqrt n}}=-2.32n​1.5​+K−1​=−2.32Plugging in K yiel: 1.65n−11.5n=1.65−n1.5=−2.32≥n=5.13\frac{\frac{1.65}{\sqrt n}-1}{\frac{1.5}{\sqrt n}}= \frac{1.65-\sqrt n}{1.5}=-2.32\geq \sqrt n=5.13n​1.5​n​1.65​−1​=1.51.65−n​​=−2.32≥n​=5.13 ≥n=26.3\geq n=26.3≥n=26.3Anbecause partiobservations are not allowe n = 27. 这道题如果没有给真实分布的话,那么可以认为type II 的概率是90%么?还是说没给真实分布的话,这道题解不出来?

2024-04-24 16:12 1 · 回答

NO.PZ2020010304000053问题如下A ta management group wants to test the null hypothesis thobserveta is N(0,1) stributeevaluating the meof a set of ranm aws. However, the actuunrlying ta is stributeN(1, 2.25). If the sample size is 10, whis the probability of a Type II error anthe power of the test? Assume a 90% confinlevel on a two-sitest.How many samples woulneeto taken to rethe probability of a Type II error to less th1%?When the null hypothesis is false, the probability of a Type II error is equto the probability ththe hypothesis fails to rejecteNow, if there are 10 samples taken from N(0,1) then the stanrviation is receH0=1/10=0.316\sigma_{H_0}=1/ \sqrt{10}=0.316σH0​​=1/10​=0.316Therefore, the cut-off points are ±1.65∗0.316=±0.522\pm1.65 * 0.316 = \pm0.522±1.65∗0.316=±0.522In actuality, the true stribution is N(1,2.25), so the σ=22.25=1.5\sigma = \sqrt{22.25} = 1.5σ=22.25​=1.5. For a sample size of 10, the expectesample stanrviation isσsample=1.5/10=0.474\sigma_{sample}=1.5/ \sqrt{10} = 0.474σsample​=1.5/10​=0.474Calculating the equivalent stanof ±1.65\pm1.65±1.65 in this stribution compareto a stanrN(0,1) yielleft = (-0.522-1)/0.474=-3.21anright =(+0.522 - 1)/ 0.474 = -1.00The probability of being on the left-hansi is practically zero. For the right, Pr( right) = 1 - Φ(-1.00) = 1 - 15.9% = 84.1%.So totprobability of a Type II error is 1 – the probability of being in the two tails isPr(Non - Rejection|Ho is false) = 1 - [Pr( left) + Pr( right)] ≈ 1 - 84.1% = 15.9%Therefore, the power of the test is 84.1%.The requirement is to have 1 - [Pr( left) + Pr( right)] = 1%Clearly, n increases from 10, the probability of being in the left-hantail will only crease from alrea being close to zero.Therefore, the requirement becomes 1 - Pr( right) = 0.01This occurs a Z-score of (using the Excel function NORMSINV) -2.32.Accorngly, the following equations neeto solve.65∗σH0=1.65/n=K1.65 * \sigma_{H_0} =1.65/ \sqrt n=K1.65∗σH0​​=1.65/n​=K an+K−11.5n=−2.32\frac{+K-1}{\frac{1.5}{\sqrt n}}=-2.32n​1.5​+K−1​=−2.32Plugging in K yiel: 1.65n−11.5n=1.65−n1.5=−2.32≥n=5.13\frac{\frac{1.65}{\sqrt n}-1}{\frac{1.5}{\sqrt n}}= \frac{1.65-\sqrt n}{1.5}=-2.32\geq \sqrt n=5.13n​1.5​n​1.65​−1​=1.51.65−n​​=−2.32≥n​=5.13 ≥n=26.3\geq n=26.3≥n=26.3Anbecause partiobservations are not allowe n = 27.老师好,这个部分使用的公式里面减去均值再除以标准差的那个X不是实际分布里面的X吗?也就是说公式是把不标准的正态分布标准化。可是这道题里面的正负0.522已经是标准正态分布里的分位点了,还可以用这个分位点的数值再标准化吗?

2024-03-06 15:19 1 · 回答