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# 統計學問題~急!!!徵高手解答

3. Allied Corporation is trying to determine whether to purchase Machine A or B. It has leased the two machines for a month. A random sample of 5 employees has been taken. These employees have gone through a training session on both machines. Below you are given information on their productivity rate on both machines. (Let the difference "d" be d = A - B.) Productivity Rate Person Machine A Machine B 1 47 52 2 53 58 3 50 47 4 55 60 5 45 53 a. State the null and alternative hypotheses for a two-tailed test. b. Find the mean and standard deviation for the difference. c. Construct a 90% confidence interval for the true difference between the machines. d. Test the null hypothesis stated in Part a at the 10% level. e. If the two machines cost the same, what would you advise Allied to do? 4. It has been suggested that night shift-workers show more variability in their output levels than day workers. Below, you are given the results of two independent random samples. Night Shift Day Shift Sample Size 9 8 Sample Mean 520 540 Sample Variance 25 23 a. State the null and alternative hypotheses to be tested. b. Compute the test statistic. c. The null hypothesis is to be tested at the 5% level of significance. State the decision rule for the test. d. What do you conclude?

### 2 Answers

- bigeyeLv 510 years agoFavorite Answer
3、 Machine A4753505545Machine B5258476053D=A-B-5-53-5-8XbarD=-4，ShatD=4.12311 a、State the null and alternative hypotheses for a two-tailed testH0：μD=0（μA=μB）H1：μD≠0（μA≠μB）b、Find the mean and standard deviation for the differenceXbarD=-4，ShatD=4.12311c、Construct a 90% confidence interval for the true difference between the machinest（0.95；5-1=4）=2.132-4-2.132*4.12311/√5≦μD≦-4+2.132*4.12311/√5-7.93122≦μD≦-0.06878d、Test the null hypothesis stated in Part a at the 10% level.t=(-4-0)/( 4.12311/√5)=-2.1693< t（0.05；5-1=4）=-2.132拒絕H0；即接受H1：μD≠0（μA≠μB）e、If the two machines cost the same, what would you advise Allied to do?XbarD=-4，D=A-B，表示Machine A生產力低於Machine B，所以建議要選Machine B 4、 Sample n Sample Mean(Xbar) Sample Variance ( Shat)Night Shift9(n1)520.00(Xbar1) 5.00( Shat1)Day Shift8(n2)540.00(Xbar2) 4.80( Shat2)依題意，兩母體變異數不相等自由度DF=（Shat1^2/n1+ Shat2^2/n2）^2/[（Shat1^2/n1）^2/(n1-1)+ （Shat2^2/n2）^2/(n2-1)]=14.893，取最接近整數值=15√（（Shat1^2/n1+ Shat2^2/n2）=2.3786t（0.975；15）=2.131a、State the null and alternative hypotheses to be tested令μ1代表夜班工人的產出均值；μ2代表日班工人的產出均值H0：μ1=μ2H1：μ1≠μ2b、Compute the test statistict=（Xbar1-Xbar2）/√（（Shat1^2/n1+ Shat2^2/n2）=-20/2.3786=-8.41c、The null hypothesis is to be tested at the 5% level of significance. State the decision rule for the testt(0.975;15)=2.131；t(0.025;15)=-2.131如果t> t(0.975;15)=2.131或t< t(0.025;15)=-2.131，拒絕H0如果t(0.025;15)=-2.131<t< t(0.975;15)=2.131，接受H0d、What do you conclude?t=-8.41< t(0.025;15)=-2.131，所以拒絕H0日夜班工人產出均值不相等，夜班工人產出水準低於日班工人產出水準

2011-07-27 17:14:36 補充：

It has been suggested that night shift-workers show more variability in their output levels than day workers.已經說變異數不同，不用檢定。

DF是以變異數不同的公式計算，算出14.893取15與變異數相同時自由度相同屬巧合。

小樣本，所以用t-test。

2011-07-27 17:17:44 補充：

這題的題意是要檢定日班和夜班工人生產水準是否不同。

- ?Lv 710 years ago
關於後一題:

(1) 題目似乎沒有說是要檢定平均數差異?

如果就字面來看, 應是檢查分散度 (標準差, 或等價地, 變異數).

然而, 若要做此檢定...

(2) 題目並無假設, 也無法推定兩班產量之分布為常態, 而樣本數又

都是那麼小...

(3) 既然假設群體變異數不等而用近似 t 檢定, 自由度為什麼取用

變異數相等之公式?