ADMS 2320 Chapter Notes -Null Hypothesis, Statistical Hypothesis Testing, Statistical Parameter
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1. The Department of Transportation would like to test the hypothesis that the average age of cars on the road is less than 10 years. A random sample of 50 cars had an average age of 9.5 years. It is believed that the population standard deviation for the age of cars is 3.4 years. The Department of Transportation would like to set = 0.10. The conclusion for this hypothesis test would be that because the test statistic is
A. more than the critical value, we can conclude that the average age of cars on the road is less than 10 years.
B. more than the critical value, we cannot conclude that the average age of cars on the road is less than 10 years.
C. less than the critical value, we can conclude that the average age of cars on the road is less than 10 years.
D. ess than the critical value, we cannot conclude that the average age of cars on the road is less than 10 years.
2. The p-value for a hypothesis test is defined as the probability of observing a
| A. critical value at least as extreme as the one selected for the hypothesis test, assuming the null hypothesis is true. | ||
| B. critical value at least as extreme as the one selected for the hypothesis test, assuming the null hypothesis is false. | ||
| C. sample mean at least as extreme as the one selected for the hypothesis test, assuming the null hypothesis is true. | ||
| D. population mean at least as extreme as the one selected for the hypothesis test, assuming the alternative hypothesis is true. |
| 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. | ||||||
Introduction: A Chi-square test is used to compare observed data with expected data according to a hypothesis. For instance, if you were crossbreeding 2 heterozygous pea plants, you would expect to see a 3:1 phenotypic ratio in the offspring. In this case, if you were to breed 400 pea plants, you would expect to see 300 plants showing the dominant trait and 100 showing the recessive trait. But what happens if you observe only 260 plants with the dominant trait and 140 plants with the recessive trait? Does this mean something is wrong with Mendelian genetics or is this difference in expected results just due to chance (random sampling error)? These are the questions that can be answered using Chi-square statistics. The results of this statistical test is used to either reject or accept (fail to reject) the null hypothesis. The null hypothesis states there is no significant difference between the observed results and the expected results. This means that if the null hypothesis is accepted, the difference in observed and expected results was just a matter of chance and so the observed results basically "fit" with what was expected. Degrees of freedom (df) = number of independent outcomes (Y) being compared less 1 df = Y-1 At the 95% confidence interval we are 95% confident that there is a significant difference between the observed and expected results, therefore rejecting the null hypothesis. Probability Value - Is the decimal value determined from the X2 table and is the probability of accepting the null hypothesis. A 0.05 probability value equates to a 95% confidence interval.
The Chi-squared test formula is: Example: If we cross two pea plants that are heterozygous yellow pods, we would expect a 3:1 phenotypic ratio. So let's say we actually did the cross and got 280 plants with green pods and 120 plants with yellow pods. Question: Is this a 3:1 phenotypic ratio? This is the value of Chi-squared Test. We have a total of 400 plants and we expect a 300 green:100 yellow phenotypic ratio If the calculated Chi-squared value is less than the critical value listed in the Chi-squared table, then we accept the null hypothesis. This means that there is no significant difference between the observed and the expected values. Our degrees of freedom (df) = 2 outcomes - 1, or df = 1. Now we go the X2 table below and using the df = 1 and probability value of 0.05, our critical value is 3.84. Since our calculated X2 value is 5.33, and is larger than the critical value, we reject the null hypothesis and can say (at 95% confidence) that there is a significant difference between our observed and expected values.
The parent generation is yellowed podded and green podded pea plants. You cross a yellow podded pea plant with a green podded pea plant and you get 100% yellow podded plants in the F1 Generation (Phenotypic ratio 4 : 0, yellow to green). What will be the expected phenotypic ratio when you allow the F1 generation to reproduce?
Fill out the Punnett square.
If we actually did the cross and got 1150 yellow and 350 green. Would this be a consistent with what was expected?
Learning Outcomes Questions
1. Why would you run a Chi-squared test?
| To determine if our data is consistent with expected results. | ||
| a To determine if our data is consistent with expected results. b To determine if our data exactly matches the expected results. | ||
| c To determine the expected results. | ||
| d | To compare the phenotypic ratios to the genotypic ratios. |
2. Determine the degrees of Freedom of the phenotypic ratio for this genetic cross.
a. 1
b. 2
c. 3
d. 4
e. 5
3. Using the data given, what is the result of your Chi-squared analysis? x2= ___.
| a. | 2.22 | |
| b | 2.71 | |
| c | 4.36 | |
| d | 187.78 | |
| e | 448.27 |
4. Using the results of your Chi-squared analysis, do we fail to reject or reject the null hypothesis?
| a. | Fail to reject the null | |
| b. | Reject the null | |
| c. | It cannot be determined from the data given |
