8 problems Note: Problems 6-8 can be solved in a separate sheet by hand, which is then scanned and include Problem 1 Can you conclude that the mean of this data set is greater than 11? Use the appropriate hypothesis test. Use an alpha of 0 ** Do this "manually" by calculating the test statistic and comparing to the critical value. You may then use built in functions 16.1138337219 12.2167904478 12.5483075554 11.7700111707 11.0881967356 6.5961246545 11.4069493795 10.4998329055 14.8886566954 11.4374398774 6.8856638461 5.838637434 13.361748833 16.6140443214 9.413674637 11.3357402172 8.853601712 15.1452545182 11.1513439172 13.5813907859 7.3712025355 6.8763537182 9.0737328172 11.6229134911 Problem 2 Can you conclude that the mean of this data set is greater than 11? Use the appropriate hypothesis test. Use an alpha of 0 ** Do this "manually" by calculating the test statistic and comparing to the critical value. You may then use built in functions 8.853601712 15.1452545182 11.1513439172 13.5813907859 7.3712025355 6.8763537182 9.0737328172 11.6229134911 Problem 3 Can you conclude that there is a real difference between X and Y? Use the appropriate hypothesis test. Use an alpha of 0.0 ** Do this "manually" by calculating the test statistic and comparing to the critical value. You may then use built in functions Problem 4 X 15.9936461183 13.0546266318 13.7458804664 17.1045855394 20.297610596 15.6317724393 15.5194582892 15.5006090978 16.9259576152 14.3025005968 9.357604379 11.9387094594 15.7111406738 13.7628952152 9.5186755695 15.9386742399 12.6515193736 11.4812328082 14.7722378093 18.0390479746 11.8589171711 12.2033808985 14.8217511387 10.5021003054 Y 16.1138337219 11.0881967356 14.8886566954 13.361748833 8.853601712 7.3712025355 12.2167904478 6.5961246545 11.4374398774 16.6140443214 15.1452545182 6.8763537182 12.5483075554 11.4069493795 6.8856638461 9.413674637 11.1513439172 9.0737328172 11.7700111707 10.4998329055 5.838637434 11.3357402172 13.5813907859 11.6229134911 Can you conclude that there is a real difference between X and Y? Use the appropriate hypothesis test. Use an alpha of 0.0 ** Do this "manually" by calculating the test statistic and comparing to the critical value. You may then use built in functions X 2.7579946999 12.5788388568 11.9328661234 15.4706101823 Y 16.1138337219 12.2167904478 12.5483075554 11.7700111707 Problem 5 Repeat P1 -P4 in Mintab. Copy the Minitab output from the session window for each of the problems below. **By doing this, you have a way of checking your work. The Minitab results should match your own calculations. Problem 6 In a process that manufactures tungsten-coated silicon wafers, the target resistance for a wafer is 85 m. In a simple random sample of 50 wafers, the sample mean resistance was 84.8 m, and the standard deviation was 0.5 m. Let represent the mean resistance of the wafers manufactured by this process. A quality engineer tests _:= := : versu _: :: . a) Find the P-value. b) Do you believe it is plausible that the mean is on target, or are you convinced that the mean is not on target? Explain your reasoning. a) Find the P-value. b) Do you believe it is plausible that the mean is on target, or are you convinced that the mean is not on target? Explain your reasoning. Problem 7 A certain manufactured product is supposed to contain 23% potassium by weight. A sample of 10 specimens of this product had an average percentage of 23.2 with a standard deviation of 0.2. If the mean percentage is found to differ from 23, the manufacturing process will be recalibrated. a) State the appropriate null and alternate hypotheses. b) Compute the P-value. c) Should the process be recalibrated? Explain. Problem 8 A process that manufactures glass sheets is supposed to be calibrated so that the mean thickness of the sheets is more than 4 mm. The standard deviation of the sheet thicknesses is known to be well approximated by = 0.20 mm. Thicknesses of each sheet in a sample of sheets will be measured, and a test of the hypothesis _: : : versus _:> :>: will be performed. Assume that, in fact, the true mean thickness is 4.04 mm. a) If 100 sheets are sampled, what is the power of a test made at the 5% level? b) How many sheets must be sampled so that a 5% level test has power 0.95? c) If 100 sheets are sampled, at what level must the test be made so that the power is 0.90? d) If 100 sheets are sampled, and the rejection region is ., what is the power ofhe test? hen scanned and included with the excel solution. esis test. Use an alpha of 0.05. Be sure to clearly state the null and alternative hypotheses. then use built in functions or the XLminer Analysis Toolpak to check your results. 9.905112875 10.5547891967 10.121299595 12.2995907008 14.4705887014 12.5920954622 15.9368342079 16.7154762581 13.9378433545 14.1189448698 11.5101286753 11.7675614271 7.7869381527 7.3174297673 10.166033969 11.1058231791 10.6726563945 12.852224225 12.4092010058 11.1622106969 8.6579483873 12.6629243264 9.6629528937 15.0577994296 12.2769130461 13.563491493 12.8524929348 14.8643177433 14.6889657001 16.5111498774 esis test. Use an alpha of 0.05. Be sure to clearly state the null and alternative hypotheses. then use built in functions or the XLminer Analysis Toolpak to check your results. 8.6579483873 12.6629243264 9.6629528937 15.0577994296 12.2769130461 13.563491493 12.8524929348 14.8643177433 14.6889657001 16.5111498774 is test. Use an alpha of 0.05. Be sure to clearly state the null and alternative hypotheses. then use built in functions or the XLminer Analysis Toolpak to check your results. 14.6794574971 15.7993897215 15.8907299225 10.2293262224 20.4813691634 16.7623755458 15.0832031483 10.3419483579 10.7840717744 13.9948803859 15.5884443482 12.5801460888 14.3910563684 12.2542263615 19.5680100139 17.2627922706 12.124411895 19.359459904 10.8090798586 15.1241587684 12.0239748519 14.6576821811 12.186240744 18.3431456534 13.8646398223 21.3955877169 14.7510752385 15.8361501302 18.775425857 16.6518218656 9.905112875 12.5920954622 11.5101286753 11.1058231791 8.6579483873 13.563491493 10.5547891967 15.9368342079 11.7675614271 10.6726563945 12.6629243264 12.8524929348 10.121299595 16.7154762581 7.7869381527 12.852224225 9.6629528937 14.8643177433 12.2995907008 13.9378433545 7.3174297673 12.4092010058 15.0577994296 14.6889657001 14.4705887014 14.1189448698 10.166033969 11.1622106969 12.2769130461 16.5111498774 is test. Use an alpha of 0.05. Be sure to clearly state the null and alternative hypotheses. then use built in functions or the XLminer Analysis Toolpak to check your results. 10.4145109013 11.2038866074 7.103587904 14.1764399789 9.7058472074 9.905112875 10.5547891967 10.121299595 12.2995907008 14.4705887014 ems below. wn calculations. fer is 85 m. In a simple eviation was 0.5 m. Let er tests _:= := : versus an is not on target? an is not on target? of 10 specimens of this centage is found to differ ness of the sheets is ximated by = 0.20 mm. is _: : : versus test? Surname 1 Student's name Professor's name Course name Date Hypothesis Testing Question 1: Testing the mean of data set is greater than 11. Null hypothesis: The mean of dataset is equal to 11. Alternative hypothesis: The mean of dataset is greater than 11. Since the sample size is greater than 30, we approximate using a normal distribution: Test statistic: z= x s/n Using hand calculation Where the sample mean is given as: 16.11383372+12.21679045+...+ 16.5111424988 X = =11.738 76268 54 Standard deviation= ( xix )2 = n1 2 2 (16.1138337211.73876268 ) + ( 12.2167904511.73876268 ) + ...+ ( 16.51 53 Surname 2 Hence the Z-statistic = 11.7387626811 =1.9658 2.761594704 / 54 Using Standardize function (Z-stat) =STANDARDIZE (AVERAGE (C8:K13), 11, STDEV.S (C8:K13)/SQRT (54)) =1.9658 This is a one-side test and the critical value at alpha =0.05 is 1.645.Since the critical value is less than test-statistic; we reject the null hypothesis and conclude that the mean is greater than 11. Question 2: Testing whether the mean is greater than 11 Null hypothesis: The mean of dataset is equal to 11. Alternative hypothesis: The mean of dataset is greater than 11. Since the sample size is less than 30, we approximate using a T-distribution: Test statistic: t= x s / n Using hand calculation sample mean ( x )= 8.853601712+15.14525452+ ...+16.51114988 =11.91526385 18 ( 8.85360171211.91526385 )2 + ( 15.1452545211.91526385 )2 +... ( 16.5111498811.91526385 )2 ==2.917703 17 standard deviation= t= 11.9152638511 =1.333 1 2.917703872/ 18 Surname 3 Using the standardize function: STANDARDIZE (AVERAGE (C22:K23), 11, STDEV.S (C22:K23)/SQRT (18)) = 1.33088751 The critical value at alpha =0.05 and n-1 =17 degrees of freedom is 1.740. Hence, we cannot reject the null hypothesis. In conclusion, the mean for dataset is 11. Question 3: Comparing two means Null hypothesis: The real mean difference between X and Y is equal to zero Alternative hypothesis: The real mean difference between X and Y is not equal to zero. Using hand calculation: Test-statistic: t= x y 0 , where SE=(s 2x /n 1)+(s2 y /n 2) SE mean ( x )=14.670904,Standard deviaton ( x ) =2.908353245; mean ( y )=11.73876268standard deviation ( y )= t= 14.67090411.738762680 2 2.908353245 2.761594704 + 54 54 2 = 2.93214132 =5.3724 0. 545773727 Using XLminer analysis toolpak: T Test: Two-Sample Assuming Unequal Variances X Y Mean 14.6709 11.73876 Variance 8.45851 7.62640 9 5 Observations 54 54 Hypothesized Mean 0 Difference Surname 4 df t Stat 105 5.37244 9 P(T<=t) one-tail 2.35E-07 t Critical one-tail 1.65949 5 P(T<=t) two-tail 4.70E-07 t Critical two-tail 1.98281 5 The two-tailed critical value is approximately 1.983.Since the critical values is less than the tstatistic we reject the null hypothesis and conclude that there exist real mean difference between datasets X and Y. Question 4: comparing two means Null hypothesis: The real mean difference between X and Y is equal to zero Alternative hypothesis: The real mean difference between X and Y is not equal to zero. Using hand calculation: Test-statistic: t= x y 0 , where SE=(s 2x /n 1)+(s2 y /n 2) SE mean ( x )=10.5938425,Standard deviaton ( x ) =3.826328196; mean ( y )=12.22225822standard deviation ( y )= t= 10.593842512.222258220 2 3.826328196 2. 033464969 + 9 9 2 = 1.62841572 =1.1274 1.444367114 Using XLminer: T Test: Two-Sample Assuming Unequal Variances X Y Surname 5 Mean Variance Observations Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail 10.5938 4 14.6407 9 9 0 12.2222 6 4.13498 9 12 -1.12743 0.14080 1 1.78228 7 0.28160 2 2.17881 3 The |test-statistic| < critical value for the two tailed test hence we do not reject the null hypothesis. In conclusion, there is no real difference between the two data sets. Question 5: Using Minitab to test Q1-Q4 For problem 1 Minitab output: One-Sample Z: Y Test of = 11 vs > 11 The assumed standard deviation = 2.76159 Variable Y N 54 Mean 11.739 StDev 2.762 SE Mean 0.376 95% Lower Bound 11.121 Z 1.97 P 0.025 SE Mean 95% Lower Bound T P For problem 2 Minitab Output: One-Sample T: W Test of = 11 vs > 11 Variable N Mean StDev Surname 6 W 18 11.915 2.918 0.688 10.719 1.33 0.100 For Problem 3 Minitab output: Two-sample T for X vs Y N 54 54 X Y Mean 14.67 11.74 StDev 2.91 2.76 SE Mean 0.40 0.38 Difference = (X) - (Y) Estimate for difference: 2.932 95% CI for difference: (1.850, 4.014) T-Test of difference = 0 (vs ): T-Value = 5.37 P-Value = 0.000 DF = 105 For problem 4 Minitab output: Two-sample T for X_1 vs Y_1 X_1 Y_1 N 9 9 Mean 10.59 12.22 StDev 3.83 2.03 SE Mean 1.3 0.68 Difference = (X_1) - (Y_1) Estimate for difference: -1.63 95% CI for difference: (-4.78, 1.52) T-Test of difference = 0 (vs ): T-Value = -1.13 P-Value = 0.282 DF = 12 All the Minitab outputs are similar to manual solution when given in two decimal places. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Question 6: In a process that manufactures tungsten-coated silicon wafers, the target resistance for a wafer is 85 m. In a simple random sample of 50 wafers, the sample mean resistance was 84.8 m, and the standard deviation was 0.5 m. Let represent the mean resistance of the wafers manufactured by this process. A quality engineer tests H 0 : =85 versus H 1 : 85 . Surname 7 a) Find the P-value. Solution: Since the sample is greater than 30, we can assume a normal distribution: Test statistic: 84.885 z= =2.828 4 0.5 50 P-value =2(P (Z<-2.8284) = 2* 0.00234 = 0.00469 b) Do you believe it is plausible that the mean is on target, or are you convinced that the mean is not on target? Explain your reasoning. Solution: No, the mean is not target because at the 0.05 level of significance we reject the null hypothesis. Question 7: A certain manufactured product is supposed to contain 23% potassium by weight. A sample of 10 specimens of this product had an average percentage of 23.2 with a standard deviation of 0.2. If the mean percentage is found to differ from 23, the manufacturing process will be recalibrated. a) State the appropriate null and alternate hypotheses. Solution: Null hypothesis: The mean percentage potassium in the product is 23. Alternative hypothesis: The mean percentage potassium in the product is not 23. b) Compute the P-value. Solution: Since the sample size is less than 30, we assume a T-distribution: 23.223 t= =3.1623 0.2 Test statistic: 10 For two-tailed test: P-value = 1-P (T< 3.1623) = 1-0.9885=0.0115 c) Should the process be recalibrated? Explain. Solution: Yes, considering a 0.05 level of significance, the null hypothesis will be rejected. In other words, mean percentage is different from 23 hence the process should be calibrated. Surname 8 Question 8: A process that manufactures glass sheets is supposed to be calibrated so that the mean thickness of the sheets is more than 4 mm. The standard deviation of the sheet thicknesses is known to be well approximated by = 0.20 mm. Thicknesses of each sheet in a sample of sheets will be measured, and a test of the hypothesis H0: 4 versus H 1 : > 4 will be performed. Assume that, in fact, the true mean thickness is 4.04 mm. a) If 100 sheets are sampled, what is the power of a test made at the 5% level? Solution: 0=100, A =4.04, =0.20, =0.05 The probability of type I error 0.05 which implies that the null hypothesis will be reject if Z 1.645 or when 0.20 x 4 +1.645 =4.0329 100 ( ) ( Power =P ( x 4.0329|=4.04 ) =P Z 4.03294.04 =P ( z 0.355 ) =1 p ( z<0.355 ) =.3594=0.64 0.2 10 ) Power of the test is 0.6406. b) How many sheets must be sampled so that a 5% level test has power 0.95? Solution: ( power=P Z ( P Z< 1.645 ( x 4.04 0.20 =0.95 , but x =4+1.645 therefore , P Z 0.2 n n ) ( ) ) 0.20 .04 1.645 ( .04 ( 0.20 ) n n ) =0.05 then =1.645, 0.2 n 0.2 n ( 0.3290.04 n=0.329, 0.04 n=0.658,n= 2 0.658 =270.6025 0.04 ) 1.645 ) .04 ( 0.20 n ) =0.95 0.2 n Surname 9 n is approximately 271. c) If 100 sheets are sampled, at what level must the test be made so that the power is 0.90? Solution: 0.20 0.20 z .04 z .04 10 10 P Z =0.9 therefore , P Z< =0.1 0.2 0.2 10 10 ( z ) ( ) ( ) ( ) .04 ( 0.20 10 ) =1.28, 0.02 z0.04=0.0256, 0.02 z =0.0144, z=0.72 0.2 10 corresponding =1P ( Z <0.72 )=10.7642=0.2358 d) If 100 sheets are sampled, and the rejection region is the test? Solution: 4.024.04 P Z =P ( Z 1 ) 0.2 10 ( ) 1P ( Z <1 )=1.1587=0.8413 the power of test is 0.8413. x 4.02 , what surname 1 student's name professor's course date hypothesis testing question 1: mean data set greater than 11. null hypothesis: dataset equal to alternative since sample size 30, we approximate using a normal distribution: statistic: z =x s n hand calculation where given as: 16.11383372+12.21679045+...+ 16.5111424988> 11 The assumed standard deviation = 2.76159 Variable Y N 54 Mean 11.739 StDev 2.762 SE Mean 0.376 95% Lower Bound 11.121 Z 1.97 P 0.025 SE Mean 95% Lower Bound T P For problem 2 Minitab Output: One-Sample T: W Test of = 11 vs > 11 Variable N Mean StDev Surname 6 W 18 11.915 2.918 0.688 10.719 1.33 0.100 For Problem 3 Minitab output: Two-sample T for X vs Y N 54 54 X Y Mean 14.67 11.74 StDev 2.91 2.76 SE Mean 0.40 0.38 Difference = (X) - (Y) Estimate for difference: 2.932 95% CI for difference: (1.850, 4.014) T-Test of difference = 0 (vs ): T-Value = 5.37 P-Value = 0.000 DF = 105 For problem 4 Minitab output: Two-sample T for X_1 vs Y_1 X_1 Y_1 N 9 9 Mean 10.59 12.22 StDev 3.83 2.03 SE Mean 1.3 0.68 Difference = (X_1) - (Y_1) Estimate for difference: -1.63 95% CI for difference: (-4.78, 1.52) T-Test of difference = 0 (vs ): T-Value = -1.13 P-Value = 0.282 DF = 12 All the Minitab outputs are similar to manual solution when given in two decimal places. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Question 6: In a process that manufactures tungsten-coated silicon wafers, the target resistance for a wafer is 85 m. In a simple random sample of 50 wafers, the sample mean resistance was 84.8 m, and the standard deviation was 0.5 m. Let represent the mean resistance of the wafers manufactured by this process. A quality engineer tests H 0 : =85 versus H 1 : 85 . Surname 7 a) Find the P-value. Solution: Since the sample is greater than 30, we can assume a normal distribution: Test statistic: 84.885 z= =2.828 4 0.5 50 P-value =2(P (Z<-2.8284) = 2* 0.00234 = 0.00469 b) Do you believe it is plausible that the mean is on target, or are you convinced that the mean is not on target? Explain your reasoning. Solution: No, the mean is not target because at the 0.05 level of significance we reject the null hypothesis. Question 7: A certain manufactured product is supposed to contain 23% potassium by weight. A sample of 10 specimens of this product had an average percentage of 23.2 with a standard deviation of 0.2. If the mean percentage is found to differ from 23, the manufacturing process will be recalibrated. a) State the appropriate null and alternate hypotheses. Solution: Null hypothesis: The mean percentage potassium in the product is 23. Alternative hypothesis: The mean percentage potassium in the product is not 23. b) Compute the P-value. Solution: Since the sample size is less than 30, we assume a T-distribution: 23.223 t= =3.1623 0.2 Test statistic: 10 For two-tailed test: P-value = 1-P (T< 3.1623) = 1-0.9885=0.0115 c) Should the process be recalibrated? Explain. Solution: Yes, considering a 0.05 level of significance, the null hypothesis will be rejected. In other words, mean percentage is different from 23 hence the process should be calibrated. Surname 8 Question 8: A process that manufactures glass sheets is supposed to be calibrated so that the mean thickness of the sheets is more than 4 mm. The standard deviation of the sheet thicknesses is known to be well approximated by = 0.20 mm. Thicknesses of each sheet in a sample of sheets will be measured, and a test of the hypothesis H0: 4 versus H 1 : > 4 will be performed. Assume that, in fact, the true mean thickness is 4.04 mm. a) If 100 sheets are sampled, what is the power of a test made at the 5% level? Solution: 0=100, A =4.04, =0.20, =0.05 The probability of type I error 0.05 which implies that the null hypothesis will be reject if Z 1.645 or when 0.20 x 4 +1.645 =4.0329 100 ( ) ( Power =P ( x 4.0329|=4.04 ) =P Z 4.03294.04 =P ( z 0.355 ) =1 p ( z<0.355 ) =.3594=0.64 0.2 10 ) Power of the test is 0.6406. b) How many sheets must be sampled so that a 5% level test has power 0.95? Solution: ( power=P Z ( P Z< 1.645 ( x 4.04 0.20 =0.95 , but x =4+1.645 therefore , P Z 0.2 n n ) ( ) ) 0.20 .04 1.645 ( .04 ( 0.20 ) n n ) =0.05 then =1.645, 0.2 n 0.2 n ( 0.3290.04 n=0.329, 0.04 n=0.658,n= 2 0.658 =270.6025 0.04 ) 1.645 ) .04 ( 0.20 n ) =0.95 0.2 n Surname 9 n is approximately 271. c) If 100 sheets are sampled, at what level must the test be made so that the power is 0.90? Solution: 0.20 0.20 z .04 z .04 10 10 P Z =0.9 therefore , P Z< =0.1 0.2 0.2 10 10 ( z ) ( ) ( ) ( ) .04 ( 0.20 10 ) =1.28, 0.02 z0.04=0.0256, 0.02 z =0.0144, z=0.72 0.2 10 corresponding =1P ( Z <0.72 )=10.7642=0.2358 d) If 100 sheets are sampled, and the rejection region is the test? Solution: 4.024.04 P Z =P ( Z 1 ) 0.2 10 ( ) 1P ( Z <1 of