第二问为什么是负的2.32呢

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

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？

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

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？

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？