MGEB12H3 Lecture Notes - Lecture 1: Simple Random Sample, Null Hypothesis, Interval Estimation
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1 | Which of the following statements about Type I and Type II errors is correct | ||||||||
a | Type I: Reject a true alternative hypothesis. Type II: Do not reject a false alternative. | ||||||||
b | Type I: Reject a true null hypothesis. Type II: Do not reject a false null hypothesis. | ||||||||
c | Type I: Reject a false null hypothesis. Type II: Reject a true null hypothesis. | ||||||||
d | Type I: Do not reject a false null hypothesis. Type II: Reject a true null hypothesis. | ||||||||
2 | You are reading a report that contains a hypothesis test you are interested in. The writer of the report writes that the p-value for the test you are interested in is 0.0749, but does not tell you the value of the test statistic. From this information you can: | ||||||||
a | Not reject the hypothesis at a Probability of Type I error = .05, but reject the hypothesis at a Probability of Type I error = 0.10 | ||||||||
b | Reject the hypothesis at a Probability of Type I error = .05, and reject at a Probability of Type I error = 0.10 | ||||||||
c | Not reject the hypothesis at a Probability of Type I error = 0.05, and not reject at a Probability of Type I error = 0.10 | ||||||||
d | Reject the hypothesis at a Probability of Type I error = .05, but not reject at a Probability of Type I error = 0.10 | ||||||||
3 | The random sample below is obtained to test the following hypothesis about the population mean. | ||||||||
H?: ? ? | 1500 | ||||||||
H?: ? > | 1500 | ||||||||
620 | 1711 | 366 | 2528 | 2678 | 1661 | 442 | 725 | 1938 | |
409 | 330 | 2480 | 542 | 369 | 2124 | 549 | 2074 | 1665 | |
1873 | 873 | 2143 | 2061 | 1177 | 2509 | 1264 | 2397 | 1523 | |
1837 | 1958 | 1041 | 1639 | 2199 | 2232 | 387 | 2270 | 2136 | |
1111 | 1883 | 2612 | 2230 | 1597 | 1726 | 694 | 1990 | 1354 | |
2090 | 909 | 2128 | 1608 | 747 | 1121 | 2220 | 2390 | 2347 | |
1041 | 316 | 655 | 632 | 2064 | 1901 | 532 | 552 | 846 | |
2704 | 1410 | 2165 | 1065 | 937 | 1452 | 2539 | 410 | 656 | |
1169 | 527 | 809 | 2364 | 2350 | 2210 | 1459 | 2391 | 856 | |
2711 | 1985 | 2382 | 2289 | 1927 | 518 | 2177 | 437 | 1151 | |
2018 | 1580 | 607 | 2715 | 2188 | 1691 | 1394 | 2610 | 1186 | |
695 | 2428 | 2246 | 858 | 2036 | 1681 | 2449 | 1578 | 1971 | |
1846 | 1729 | 2389 | 1737 | 1913 | 1863 | 2072 | 2593 | 2287 | |
2220 | 2230 | 551 | 458 | 2626 | 2731 | 488 | 2551 | 1736 | |
1373 | 307 | 1803 | 2647 | 2679 | 1508 | 1468 | 1443 | 516 | |
1002 | 2116 | 2616 | 817 | 2522 | 460 | 1879 | 1999 | 1837 | |
The level of significance of the test is ? = 0.05. Compute the relevant test statistic. | |||||||||
This is a(n) _______ (two-tail, upper-tail, lower-tail) test. The test statistic is TS = _______. | |||||||||
a | Two-tail test | TS = | 1.81 | ||||||
Do not reject H?: ? ? 1500. Conclude that the population mean is not greater than 1500. | |||||||||
b | Upper tail test. | TS = | 1.52 | ||||||
Do not reject H?: ? ? 1500. Conclude that the population mean is not greater than 1500. | |||||||||
c | Upper tail test. | TS = | 1.81 | ||||||
Reject H?: ? ? 1500. Conclude that the population mean is greater than 1500. | |||||||||
d | Lower tail test. | TS = | 1.98 | ||||||
Do not reject H?: ? ? 1500. Conclude that the population mean is no greater than 1500. | |||||||||
4 | Consider the following hypothesis test. | ||||||||
H?: ? ? | 30 | ||||||||
H?: ? > | 30 | ||||||||
A random sample of n = 15 yielded the following observations | |||||||||
51 | 38 | 26 | 16 | 28 | |||||
57 | 20 | 33 | 35 | 23 | |||||
21 | 47 | 56 | 54 | 36 | |||||
Use ? = | 0.05 | ||||||||
TS = ______ | CV = ______ | State the decision rule. | |||||||
a | 1.68 | 1.761 | Do not reject H?. Conclude the mean is not greater than 30. | ||||||
b | 1.68 | 1.64 | Reject H?. Conclude the mean is greater than 30. | ||||||
c | 1.847 | 2.145 | Do not reject H?. Conclude the mean is not less than 30. | ||||||
d | 1.847 | 1.761 | Reject H?. Conclude the mean is less than 30. | ||||||
5 | In a recent study, a major fast food restaurant had a mean service time of 165 seconds. The company embarks on a quality improvement effort to reduce the service time and has developed improvements to the service process. The new process will be tested in a sample of stores. The new process will be adopted in all of its stores, if it reduced mean service time by more than 45 seconds compared to the current mean service time. To perform the hypothesis test, the sample of 48 stores yields the following data (seconds). | ||||||||
90 | 96 | 133 | 108 | 136 | 110 | 119 | 138 | ||
129 | 98 | 101 | 92 | 135 | 124 | 115 | 90 | ||
132 | 125 | 110 | 124 | 126 | 138 | 94 | 130 | ||
108 | 96 | 140 | 135 | 102 | 114 | 109 | 137 | ||
138 | 104 | 108 | 134 | 92 | 107 | 96 | 119 | ||
105 | 111 | 96 | 136 | 126 | 116 | 98 | 131 | ||
Use ? = | 0.05 | ||||||||
|TS| = ______ | |CV| = ______ | ||||||||
a | 1.548 | 1.678 | Do not reject H?. The mean service time is not reduced by more than 45 seconds. Do not adopt the new process. | ||||||
b | 1.871 | 1.678 | Reject H?. The mean service time is reduced by more than 45 seconds. Adopt the new process. | ||||||
c | 1.871 | 1.640 | Do not reject H?. The mean service time is not reduced by more than 45 seconds. Do not adopt the new process. | ||||||
d | 1.548 | 1.640 | Reject H?. The mean service time is reduced by more than 45 seconds. Adopt the new process. | ||||||
1. You are given only three quarterly seasonal indices and quarterly seasonally adjusted data for the entire year. What is the raw data value for Q4? Raw data is not adjusted for seasonality.
Quarter Seasonal Index Seasonally Adjusted Data
Q1 .80 295
Q2 .85 299
Q3 1.15 270
Q4 --- 271
2. One model of exponential smoothing will provide almost the same forecast as a liner trend method. What are linear trend intercept and slope counterparts for exponential smoothing?
A. Alpha and Delta
B. Delta and Gamma
C. Alpha and Gamma
D. Standard Deviation and Mean
3. When performing correlation analysis what is the null hypothesis? What measure in Minitab is used to test it and to be 95% confident in the significance of correlation coefficient.
A. Ho: r = .05 p < .5
B. Ho: r = 0 p >.05
C. Ho: r ? 0 p?.05
D. Ho: r = 0 p?.05
In decomposition what does the cycle factor (CF) of .80 represent for a monthly forecast estimate of a Y variable? |
A. The estimated value is 80% of the average monthly seasonal estimate.
B. The estimate is .80 of the forecasted Y trend value.
C. The estimated value is .80 of the historical average CMA values.
D. The estimated value has 20% more variation than the average historical Y data values.
5. A Wendy's franchise owner notes that the sales per store has fallen below the stated national Wendy's outlet average of $1,368,000. He asserts a change has occurred that reduced the fast food eating habits of Americans. What is his hypothesis (H1) and what type of test for significance must be applied? |
A. H1: u ? $1,368,000 A one-tailed t-test to the left.
B. H1: u = $1,368,000 A two-tailed t-test.
C. H1: u < $1,368,000 A one-tailed t-test to the left.
D. H1: p < $1,368,000 A one-tailed test to the right
A. The rejection region and the t-table value generally gets smaller for sample size below 31. |
A. Yes. The data are significantly correlated through the 12th lag. C. No. Only the 12 lag period is not correlated. D. You cannot tell since the number of sample observations is not provided. E. The p-value is above .05 so the data is correlated. |
A. Type 2 error |
A. Yes. They move in the same direction as statistical significance. |
A. The weight cannot be calculated since the data observation is not given. |
A. Yes. The correlation coefficient is .873 that is greater than .05. |
A. Yes, since the residuals randomly vary in magnitude. |
A. -101.0 |
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