(8 Questions) 1. The mean weight of the adults in a certain region is believed to be . An agency thinks that this mean may be higher than the above belief. They are going to take a simple random sample of 100 adults for a hypothesis test at 5% level of significance. a) State the null and the alternate hypothesis for this test. b) If the mean weight of the adults in the above simple random sample is and we know that the population standard deviation is , compute the test statistic and the P_value of the test. c) Will you reject or not reject the null hypothesis at 5% level of significance? 2. An expensive medication from a particular company states 100 ml for the contents on the bottles for this medication. An agency would like to test if the bottles are being under filled, in the sense that the mean contents of all bottles is less than 100 ml. a) State the null hypothesis and the alternative hypothesis. b) The contents in 10 randomly selected bottles were measured, and the results are below in ml. contents in ml 96.0 102.0 99.4 100.5 98.5 99.4 Variable contents N 10 Atul N Roy N* 0 Mean 99.300 SE Mean 0.696 97.5 StDev 2.201 98.7 103.4 Minimum 96.000 Q1 97.575 Median 99.050 97.6 Q3 100.875 Page 1 Variable contents Maximum 103.400 State the test statistic State the P_value. Does this sample show a good evidence at 5% level of significance that the bottles are being under filled? 3. The electricity supply in a certain region appears to show a lot of voltage fluctuation from the required value of 240 V. To check if the mean voltage is different from 240 V, the voltage will be measured at 15 randomly selected times. a) State the null and the alternative hypothesis for this test. b) The following are the readings shown by the 15 measurements. voltage 222.2 217.9 211.9 219.7 208.8 229.5 219.4 208.1 209.0 208.2 196.2 211.7 199.9 201.1 218.0 Variable voltage N 15 N* 0 Variable voltage Maximum 229.50 Mean 212.11 SE Mean 2.35 StDev 9.10 Minimum 196.20 Q1 208.10 Median 211.70 Q3 219.40 a) Write the test statistic. b) Will you reject or not reject the null hypothesis at 5% level of significance? c) Compute a 95% confidence interval for the overall mean voltage based on the results from the above sample. Atul N Roy Page 2 d) Does this interval contain the value 240 Volt? e) Explain the usage of a 95% confidence interval for a two sided hypothesis test for a population mean at a 5% level of significance. 4. In the past, 44% of those taking a public accounting qualifying exam have passed the exam on their first try. Lately, the availability of exam preparation books and tutoring sessions may have improved the likelihood of an individual's passing on his or her first try. To test if the likelihood of passing in the first try in now higher, a simple random sample of recent applicants will be examined. a) State the null hypothesis and the alternative hypothesis b) In a sample of 250 recent applicants, 130 passed on their first attempt. At 0.05 level of significance, can we conclude that the proportion passing on the first try has increased? State the P_value and a reason for your conclusion. 5. We are testing at 1% level of significance for the presence of an attribute in two populations 1 and 2. The results from independent simple random samples are shown below. Atul N Roy Page 3 Population 1 Population 2 # with the attribute 56 198 Sample size 132 489 a) State the test statistic, the P_value, and your conclusion about rejecting or not rejecting the null hypothesis. b) Compute a 99% confidence interval for p1 p2 and describe whether the information from this interval is consistent with the conclusion in the hypothesis test. 6. Explain the meaning of a type I error and a type II error. 7. We are going to test the null and the alternative hypotheses at a 1% level of significance H 0 : 80 H A : 80 based on a simple random sample of size 50 and the population standard deviation 20 . a) Compute the probability of a type II error if actually 75 Atul N Roy Page 4 b) Compute the power of the test if actually 72 c) Compute the probability of a type II error if actually 70 d) Compute the power of the test if actually Atul N Roy 65 Page 5 8. The following table shows the # of ice cream bars sold by a vendor as explained by the outside high temperature High Temperature for the day in degress F 59 68 # of bars of ice cream sold 11 32 a) b) c) d) e) f) 79 86 97 73 76 151 SS SS Find SS Write the equation of the line of best fit (the least square regression line) Use the line of best fit to predict the # of cups sold on a day with a high temperature of 70 degrees F. What is the value of the correlation coefficient between the two variables. What percentage of variation in the # of cups sold is explained by the line of best fit? Is the correlation significant at 5% level of significance? Atul N Roy xx xy yy Page 6 Quiz 3 (8 Questions) 1. The mean weight of the adults in a certain region is believed to be . An agency thinks that this mean may be higher than the above belief. They are going to take a simple random sample of 100 adults for a hypothesis test at 5% level of significance. a) State the null and the alternate hypothesis for this test. Null hypothesis: mean=160lb Alternative hypothesis: mean < 160lb b) If the mean weight of the adults in the above simple random sample is and we know that the population standard deviation is , compute the test statistic and the P_value of the test. (173.53-160)/(35/10)=P value=3.87 c) Will you reject or not reject the null hypothesis at 5% level of significance? I will reject the null hypothesis at 5% level of significance. 2. An expensive medication from a particular company states 100 ml for the contents on the bottles for this medication. An agency would like to test if the bottles are being under filled, in the sense that the mean contents of all bottles is less than 100 ml. a) State the null hypothesis and the alternative hypothesis. Null hypothesis: mean=100ml Alternative hypothesis: mean<100ml b) the contents in 10 randomly selected bottles were measured, and results are below ml. ml 96.0 102.0 99.4 100.5 variable n n* q3 atul roy 97.5 98.7 103.4 97.6 mean se stdev minimum 98.5 q1 median page 1 0 99.300 0.696 2.201 96.000 97.575 99.050 100.875 maximum 103.400 state test statistic 99.300-100>0.44 b) In a sample of 250 recent applicants, 130 passed on their first attempt. At 0.05 level of significance, can we conclude that the proportion passing on the first try has increased? State the P_value and a reason for your conclusion. P value= 0.52 Because 0.0054<0.05, we reject the null hypothesis and conclude that the population portion is greater than 0.44. 5. We are testing Atul N Roy Page 3 H o : p1 p2 H A : p1 p2 at 1% level of significance for the presence of an attribute in two populations 1 and 2. The results from independent simple random samples are shown below. Population 1 Population 2 # with the attribute 56 198 Sample size 132 489 a) State the test statistic, the P_value, and your conclusion about rejecting or not rejecting the null hypothesis. Statistic= (0.424-0.405)/ (0.424 (0.576)/132 +0.405*0.595/489 =0.019/0.048404 =0.3925 The p value from the tables =0.39811 The p-value is greater than 0.05, 0.10, and 0.001 Therefore we do not reject the null hypotheses. We therefore conclude that there is no difference. Hence no attribute. b) Compute a 99% confidence interval for p1 p2 and describe whether the information from this interval is consistent with the conclusion in the hypothesis test. Difference = (0.424-0.405)2.58*0.048404 =0.0190.125=(-0.107,0.145) Atul N Roy Page 4 Calculated t statistic 0.3925 is outside the interval we there conclude that the results are consistent. 6. Explain the meaning of a type I error and a type II error. Type I error is when the null hypothesis is rejected. Type II error is when we fail to reject a false hypothesis. 7. We are going to test the null and the alternative hypotheses at a 1% level of significance H 0 : 80 H A : 80 based on a simple random sample of size 50 and the population standard deviation 20 . a) Compute the probability of a type II error if actually 75 = (75-80)*50/20 =-1.767766 P=1-0.21186 =0.78814 b) 72 Compute the power of the test if actually Atul N Roy Page 5 = (72-80)*50/20 =-2.828 P=1-0.3446=0.6554 Compute the probability of a type II error if actually 70 = (70-80)*50/20 =-3.536 P=1-0.3085=0.6915 c) Compute the power of the test if actually 65 = (65-80)*50/20 =5.303 P=1-0.22663=0.77337 Atul N Roy Page 6 8. The following table shows the # of ice cream bars sold by a vendor as explained by the outside high temperature High Temperature for the day in degress F 59 # of bars of ice cream sold 11 a) 68 32 79 73 Write the equation of the line of best fit (the least square regression line) Linear equation will be Y=-202.454 +3.484X b) Use the line of best fit to predict the # of cups sold on a day with a high temperature of 70 degrees F. Ice cream sales= -202.454+3.484 temperature Ice cream sales= -202.454+3.484(70) =-202.454+243=41.426 c) What is the value of the correlation coefficient between the two variables. =0.09668 d) What percentage of variation in the # of cups sold is explained by the line of best fit? This is explained by the value of r squared 0.096682= 0.00934 Atul N Roy Page 7 86 97 76 151 Quiz 3 (8 Questions) 1. The mean weight of the adults in a certain region is believed to be . An agency thinks that this mean may be higher than the above belief. They are going to take a simple random sample of 100 adults for a hypothesis test at 5% level of significance. a) State the null and the alternate hypothesis for this test. Null hypothesis: mean=160lb Alternative hypothesis: mean < 160lb b) If the mean weight of the adults in the above simple random sample is and we know that the population standard deviation is , compute the test statistic and the P_value of the test. (173.53-160)/(35/10)=P value=3.87 c) Will you reject or not reject the null hypothesis at 5% level of significance? I will reject the null hypothesis at 5% level of significance. 2. An expensive medication from a particular company states 100 ml for the contents on the bottles for this medication. An agency would like to test if the bottles are being under filled, in the sense that the mean contents of all bottles is less than 100 ml. a) State the null hypothesis and the alternative hypothesis. Null hypothesis: mean=100ml Alternative hypothesis: mean<100ml b) the contents in 10 randomly selected bottles were measured, and results are below ml. ml 96.0 102.0 99.4 100.5 variable n n* q3 atul roy 97.5 98.7 103.4 97.6 mean se stdev minimum 98.5 q1 median page 1 0 99.300 0.696 2.201 96.000 97.575 99.050 100.875 maximum 103.400 state test statistic 99.300-100>0.44 b) In a sample of 250 recent applicants, 130 passed on their first attempt. At 0.05 level of significance, can we conclude that the proportion passing on the first try has increased? State the P_value and a reason for your conclusion. P value= 0.52 Because 0.0054<0.05, we reject the null hypothesis and conclude that the population portion is greater than 0.44. 5. We are testing Atul N Roy Page 3 H o : p1 p2 H A : p1 p2 at 1% level of significance for the presence of an attribute in two populations 1 and 2. The results from independent simple random samples are shown below. Population 1 Population 2 # with the attribute 56 198 Sample size 132 489 a) State the test statistic, the P_value, and your conclusion about rejecting or not rejecting the null hypothesis. Statistic= (0.424-0.405)/ (0.424 (0.576)/132 +0.405*0.595/489 =0.019/0.048404 =0.3925 The p value from the tables =0.39811 The p-value is greater than 0.05, 0.10, and 0.001 Therefore we do not reject the null hypotheses. We therefore conclude that there is no difference. Hence no attribute. b) Compute a 99% confidence interval for p1 p2 and describe whether the information from this interval is consistent with the conclusion in the hypothesis test. Difference = (0.424-0.405)2.58*0.048404 =0.0190.125=(-0.107,0.145) Atul N Roy Page 4 Calculated t statistic 0.3925 is outside the interval we there conclude that the results are consistent. 6. Explain the meaning of a type I error and a type II error. Type I error is when the null hypothesis is rejected. Type II error is when we fail to reject a false hypothesis. 7. We are going to test the null and the alternative hypotheses at a 1% level of significance H 0 : 80 H A : 80 based on a simple random sample of size 50 and the population standard deviation 20 . a) Compute the probability of a type II error if actually 75 = (75-80)*50/20 =-1.767766 P=1-0.21186 =0.78814 b) 72 Compute the power of the test if actually Atul N Roy Page 5 = (72-80)*50/20 =-2.828 P=1-0.3446=0.6554 Compute the probability of a type II error if actually 70 = (70-80)*50/20 =-3.536 P=1-0.3085=0.6915 c) Compute the power of the test if actually 65 = (65-80)*50/20 =5.303 P=1-0.22663=0.77337 Atul N Roy Page 6 8. The following table shows the # of ice cream bars sold by a vendor as explained by the outside high temperature High Temperature for the day in degress F 59 # of bars of ice cream sold 11 a) 68 32 79 73 Write the equation of the line of best fit (the least square regression line) Linear equation will be Y=-202.454 +3.484X b) Use the line of best fit to predict the # of cups sold on a day with a high temperature of 70 degrees F. Ice cream sales= -202.454+3.484 temperature Ice cream sales= -202.454+3.484(70) =-202.454+243=41.426 c) What is the value of the correlation coefficient between the two variables. =0.09668 d) What percentage of variation in the # of cups sold is explained by the line of best fit? This is explained by the value of r squared 0.096682= 0.00934 Atul N Roy Page 7 86 97 76 151