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# HW6-1 - Ofﬂine HW#6 I The diameter of 3.5 inch diskettes...

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Unformatted text preview: Ofﬂine HW #6 I. The diameter of 3.5 inch diskettes is normally distributed. Periodically, quality control inspectors randomly select a sample of 23 diskettes. If the mean diameter of the diskettes is too large or too small the diskette punch is shut down for adjustment; otherwise, the punching process continues. The last sample showed a mean and standard deviation of 3.47 and 0.08 inches, respectively. Using a = 0.01, the appropriate decision is : do not reject the null hypothesis and do not shut down the punch reject the null hypothesis and shut down the punch reject the null hypothesis and do not shut down the punch do not reject the null hypothesis and shut down the punch 1, The diameter of 3.5 inch diskettes is normally distributed. Periodically, quality control inspectors randomly select a sample of 18 diskettes. If the mean diameter of the diskettes is too large or too small the diskette punch is shut down for adjustment; otherwise, the punching process continues. The last sample showed a mean and standard deviation of 3.55 and 0.08 inches, respectively. Using a = 0.05, the appropriate decision is: do not reject the null hypothesis and do not shut down the punch reject the null hypothesis and do not shut down the punch do not reject the null hypothesis and shut down the punch reject the null hypothesis and shut down the punch 3, A market researcher is testing consumer opinions about a new product. Consumers are asked to rate the product on a scale of from 0 to 10 as to whether or not they would purchase the product if it were available (where 0 = absolutely would not purchase, 10 = absolutely would purchase). Past studies have shown that this scale is normally distributed in the population. The market researcher will not give the go ahead for further development of the product unless the population mean purchase score is greater than 6.00. A sample of eight scores are obtained as follows: 7, 8, 5, 6, 9, 7, 5, 8. The observed t statistic is: 1.81 4.83 1.70 0.21 ‘l’ A company believes that it controls more than 40% of the total market share for one of its products. To prove this belief, a random sample of 165 purchases of this product are contacted. It is found that 70 of the 165 purchased this company's brand of the product. If a researcher wants to conduct a statistical test for this problem, the altemative hypothesis would be: the population proportion is less than 0.40. the population proportion is greater than 0.40. the population mean is equal to 0.40. the population proportion is not equal to 0.40. A company believes that it controls more than 40% of the total market share for one of its products. To prove this belief, a random sample of 16S purchases of this product is taken. It is found that 70 of the 165 purchased this company's brand of the product. If a researcher wants to conduct a statistical test for this problem, the observed 2 value is: 0.63 1.03 0.35 1.24 G, A bank inspector monitors the default rate on personal loans at the AFB member banks. One standard that she examines is that no more than 5% of personal loans should be in default. On each Friday, the default rate is calculated for a sample of 500 personal loans. Last Friday’s sample contained 30 defaulted loans. The bank inspector's alternate hypothesis is: p < 0.05 p=.05 p¢.05 p > 0.05 *7. A bank inspector monitors the default rate on personal loans at the AFB member banks. One standard that she examines is that no more than 5% of personal loans should be in default. On each Friday, the default rate is calculated for a sample of 500 personal loans. Last Friday's sample contained 30 defaulted loans. If the bank inspector uses hypothesis testing techniques to test this and if alpha is .10, then the critical value is: 1.645 -1.28 -1.645 1.28 8' A bank inspector monitors the default rate on personal loans at the AFB member banks. One standard that she examines is that no more than 5% of personal loans should be in default. On each Friday, the default rate is calculated for a sample of 500 personal loans. Last Fn'day's sample contained 30 defaulted loans. If the bank inspector uses hypothesis testing techniques to test this and if alpha is .10, then the observed value is: 0.046 -1.03 1.03 -0.045 ‘1. Management at a large company has recently implemented a more participatory management style. Management believes that at least half of the company's employees feel that they are more a part of the decision making process than before. To test this, the company’s business researcher randomly polls 180 company employees. One hundred and seven of those polled say that they feel they are more a part of the decision making process than before. Suppose the researcher is using a 1% level of significance. The critical value is: 1.645 1.96 2.33 2.575 m. Management at a large company has recently implemented a more participatory management style. Management believes that at least half of the company's employees feel that they are more a part of the decision making process than before. To test this, the company's business researcher randomly polls 180 company employees. One hundred and seven of those polled say that they feel they are more a part of the decision making process than before. Suppose the researcher is using a 1% level of significance. The observed value is: 2.53 1.84 2.67 1.56 H . A graphical curve constructed by plotting beta values against various values of the alternative hypothesis is called: Pareto chart power curve operating characteristic curve ogive curve 57.. The following data (in pounds), which were selected randomly from a normally distributed population of values, represent measurements of a machine part that is supposed to weigh, on average, 8.3 pounds. 8.1 8.4 8.3 8.2 8.5 8.6 8.4 8.3 8.4 8.2 8.8 8.2 8.2 8.3 8.1 8.3 8.4 8.5 8.5 8.7 A110 THC mums ou'mn' Use these data and a : .01 to test the hypothesis that the parts average 8.3 pounds. USE 3 R9994) RCHES ‘. THE 1557' STATISTK’THG \$‘VN.V€,AK!) THE (Ir il Test of mu 8.3 vs not = 8.3 Variable N Mean StDev SE Mean 99% CI '1‘ P Weights 20 8.3700 0.1895 0.0424 (8.2488, 8.4912) 1.65 0.115 I3- One survey conducted by RHI Management Resources determined that the Lexus is the favorite luxury car for 25% of CFOs. Suppose a financial management association conducts its own survey of CFOs in an effort to determine whether this figure is correct. They use an alpha of .05. Following is the MINITAB output with the results of the survey. Discuss the ﬁndings, including the hypotheses, one— or two-tailed tests, sample statistics, and the conclusion. Explain from the data why you reached the conclusion you did. . " “8‘ 3 «WRWCHU ‘Q ‘i‘ssmrzé(“1fmiirsw Wawrtz’en mt ‘ “KB A C (“03‘6“ ‘ Test of p z 0.28 vs 39 not 2 0.25 ° 11“ 7‘“ STM’ih‘V. Exact Sample X 3:“ Sample p 953’. 1‘31 P—‘i'alue ‘ TM 1' Name 1 .79 394 0.2052729to.1663§9,0.24§61533 0.04s '1‘“: (I: 9‘4 . A researcher wants to conduct a before/after study on 15 subjects to determine if a treatment results in any difference in scores. The null hypothesis is that the average difference is zero while the alternative hypothesis is that the average difference is not zero. Scores are obtained on the subjects both before and after the treatment. After subtracting the after scores from the before scores, the average difference is computed to be ~2.13 with a sample standard deviation of 1.21. The degrees of freedom for this test are 28 15 14 13 is. A researcher wants to conduct a before/after study on 15 subjects to determine if a treatment results in any difference in scores. The null hypothesis is that the average difference is zero while the alternative hypothesis is that the average difference is not zero. Scores are obtained on the subjects both before and after the treatment. After subtracting the after scores from the before scores, the average difference is computed to be —2.13 with a sample standard deviation of 2.51. The observed t value for this test is: -3.29 -1.76 -2.87 -3.04 n. . A researcher wants to conduct a before/after study on 15 subjects to determine if a treatment results in any difference in scores. The null hypothesis is that the average difference is zero while the altemative hypothesis is that the average difference is not zero. Scores are obtained on the subjects both before and after the treatment. After subtracting the after scores from the before scores, the average difference is computed to be -2.13 with a sample standard deviation of 1.21. Using an alpha of .05, the table t value for this test is: 2.131 2.145 2.262 1.761 ...
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HW6-1 - Ofﬂine HW#6 I The diameter of 3.5 inch diskettes...

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