Unformatted text preview: Math 241Exam 3 Practice Problems This exam will cover chapters 7‐9. The major focus will be on hypothesis testing (Z‐Test & T‐Test, including the critical value and p‐value approaches). The Central Limit Theorem is also a very important idea. A brief summary for hypothesis testing follows below. Suppose we are given a normally distributed population OR a large sample of 30 or greater (simple random sample). Use a Z‐Test if: •
The standard deviation of the POPULATION is given. Use a T‐Test if: •
The standard deviation of the SAMPLE is given. For either test, you may use: •
A critical‐value approach (i.e. obtain a critical value and compare with the test‐statistic); •
A p‐value approach (the probability the sample mean occurs given the null hypothesis is true; compare this value with the level of significance). If p < alpha, REJECT the null hypothesis; if p > alpha, FAIL TO REJECT! 1. Mean Household Income According to reports, the average household income in Fairfax County is $105,000. Highly skeptical of this claim, you believe the actual household income is different from this value. To test this claim, a simple random sample of 36 households in Fairfax generated a mean income of $95,000. The population standard deviation is $25,000. What are the null and alternative hypotheses? Which statistical test is appropriate: Z‐Test or T‐Test? Is this a one‐tailed or two‐tailed test? What is (are) the critical value(s)? What is the test statistic? Using the critical‐value approach, test the null hypothesis at the 5% level of significance. Sketch a normal curve labeling the “population mean,” critical value, and test statistic. Shade the rejection region. Using the p‐value approach, test the null hypothesis at the 5% level of significance. Using the critical‐value approach, test the null hypothesis at the 1% level of significance. 2. Meat Consumption! According to the U.S Department of Agriculture, the average American consumes a half pound of meat per day. Since we believe people eat more than this, we test the claim with a random sample of 16 individuals. The sample mean meat consumption is .55 pounds with a sample standard deviation of .1 pounds. Assume the population is normally distributed. What are the null and alternative hypotheses? Which statistical test is appropriate: Z‐Test or T‐Test? Is this a one‐tailed or two‐tailed test? What is (are) the critical value(s)? What is the test statistic? Using the critical‐value approach, test the null hypothesis at the 5% level of significance. Sketch a normal curve labeling the “population mean,” critical value, and test statistic. Shade the rejection region. Using the p‐value approach, test the null hypothesis at the 5% level of significance. Using the critical‐value approach, test the null hypothesis at the 1% level of significance. 1 3. Get your sleep! A group of psychologists claim that the average student sleep less than 5 hours the night before a morning exam. We sample 29 students who each had an exam the next morning. The sample mean is 5.2 hours with a sample standard deviation of 15 minutes. Test the claim at the .05 level of significance. Assume the population is normally distributed. • : 5; : 5 • T‐Test since sample standard deviation is given. • Critical Value = 1.701 • Test Statistic = 4.31 • This is a one‐tailed (right‐tail) test! • P‐Value < .001 Since 4.31 > 1.701, REJECT the null hypothesis. We have sufficient evidence to conclude that the mean sleep for students before an exam is GREATER THAN OR EQUAL TO 5 hours. Alternatively, note that p < .05, so reject (small p‐value). 4. Average Test Grade A student believes that the average grade on the statistics final exam is 87. A sample of 10 final exams is taken from a large class. The average grade in the sample is 83.96. The population standard deviation is 12. Test the claim at the .05 level of significance. Assume the population of scores is normally distributed. • : 87; : 87 • This is a two‐tailed test! • T‐Test since sample standard deviation is given. • Critical Value = 2.262 & ‐2.262 • Test Statistic = ‐.90 • P‐Value = .39 Since ‐0.90 > ‐2.262, we FAIL TO REJECT the null hypothesis. The student DOES NOT have sufficient evidence to conclude that the average test grade is 87. Alternatively, note that .39 > .05, so FAIL TO REJECT (p‐value too large). 5. Carpet Delivery A carpet company advertises that it will deliver your carpet in exactly 15 days of the purchase. A sample of 49 past customers is taken. The average delivery time in the sample was 16.2 days. The population standard deviation is 5.6 days. Test the claim at the .10 level of significance. • : 15; : 15 • This is a two‐tailed test! • Z‐Test since population standard deviation is given. • Critical Values = 1.645 & ‐1.645 • Test Statistic = 1.5 • P‐Value = .1336 Since 1.5 < 1.645, we FAIL TO REJECT the null hypothesis. We do not have sufficient evidence to reject the carpet company’s claim. Alternatively, note that .1336 > .10, so FAIL TO REJECT (p‐value too large). 2 6. Raising a Child The U.S. department of agriculture reports the mean cost of raising a child from birth to age 5 in a rural area is $15,000. You believe this value is inaccurate, so you select a random sample of 1,000 children (all age 5) and find the mean cost is $15,500. The population standard deviation is $1,500. At the .05 level of significance, is there enough evidence to conclude that the mean cost is more than $15,000? • : 15; : 15 • This is a one‐tailed test (right‐tailed)! • Z‐Test since population standard deviation is given. • Critical Value = 1.645 • Test Statistic = 10.5 • P‐Value < .001 Since 10.5 > 1.645, we REJECT the null hypothesis. We have sufficient evidence to reject the department’s claim that the average cost for raising a child from birth to age 5 is $15,000. Alternatively, note that p‐value < .05, so REJECT (small p‐value). 7. The mean and standard deviation of a population are, respectively, 156.2 and 40.5. What are the mean and standard deviation of the sample mean with n 64 ? 8. The weight of a male’s brain is normally distributed with mean 1,325 grams. Suppose we randomly select simple random samples of 29 males each and weigh their brains. What can we conclude from the Central Limit Theorem? 9. The salaries for NBA players are not known to be normally distributed. Suppose we select 1,000 simple random samples of 45 NBA players in each sample and record the sample mean salaries (so we will have 1,000 sample means). What can we conclude from the Central Limit Theorem? 10. The mean rent for an apartment in Sterling is $1,500. You randomly select nine of these apartments. Assume that the rents are normally distributed, with a standard deviation of $600. Describe the distribution of sample means. What is the probability that the sample mean rent (for a random sample of size 9) is less than $1800? 11. A statistics exam takes an average of 45 minutes to be completed. The standard deviation is 10 minutes. A random sample of 49 students is selected. What is the probability that a student takes less than 44 minutes to finish? What is the probability that a sample mean of 49 students is less than 44 minutes? 12. For a t‐curve with 15 degrees of freedom, find . . 13. If we conduct a t‐test and obtain a test‐statistic from a sample of size 15, what will be the corresponding degrees of freedom? 14. A hypothesis test is to be performed for a population mean with null hypothesis 0 . The population is normally distributed and the test statistic used will be ⁄√ The store manager of Giant claims that the mean amount of time customers spend waiting in line is less than 4.2 minutes. You wish to test this claim at the 0.05 level of significance. The mean waiting time for a random sample of n 81 customers is 3.0 minutes. Assume 1.0. Compute the value of the test statistic. 15. Suppose you wish to conduct a hypothesis test for a population mean with the following known information: 1 The population standard deviation is known, 2 the population is far from being normally distributed, 3 the sample size is large and 4 simple random sampling is utilized. Should you use a Z‐Test or T‐Test? 3 16. Suppose you wish to conduct a hypothesis test for a population mean with the following known information: 1 The sample standard deviation is known, 2 there are outliers present, 3 the sample size is medium and 4 simple random sampling is utilized. Should you use a Z‐Test or T‐Test? 17. Suppose you wish to conduct a hypothesis test for a population mean with the following known information: 1 The population standard deviation is known, 2 the population is normally distributed, 3 the sample size is small and 4 convenience sampling is utilized. Should you use a Z‐Test or T‐Test? 18. What is the primary purpose for conducting hypothesis testing [1‐sample Z‐Tests and T‐Tests]? 19. What is a critical value? 20. Which NBA player leads all centers for the highest scoring average per game in the playoffs? _____ 21. Suppose we conduct a one‐tailed hypothesis test (with the rejection region in the left tail) with a critical value of ‐
1.87. Further, suppose the test statistic is ‐1.97. What can we conclude? We would fail to reject the null hypothesis. We would reject the null hypothesis in favor of the alternative hypothesis. Not enough information is provided. _____ 22. Suppose we conduct a one‐tailed hypothesis test (with the rejection region in the right tail) with a critical value of 1.87. Further, suppose the test statistic is 1.70. What can we conclude? We would fail to reject the null hypothesis. We would reject the null hypothesis in favor of the alternative hypothesis. Not enough information is provided. _____ 23. Suppose we conduct a one‐tailed hypothesis test (with the rejection region in the right tail) with a test statistic of 2.0. Thus, the p‐value is roughly .025 normalcdf 2.0, 10000 . Assume a 5% level of significance. What can we conclude? We would fail to reject the null hypothesis. We would reject the null hypothesis in favor of the alternative hypothesis. Not enough information is provided. _____ 24. Suppose we conduct a hypothesis test and the p‐value is less than .001. What can we conclude? We would fail to reject the null hypothesis. We would reject the null hypothesis in favor of the alternative hypothesis. Not enough information is provided. _____ 25. Consider a study which investigates the average number of hours an individual watches television in a given day. According to Communications Industry Forecast & Report, the average person watches 4.66 hours of television [null hypothesis]. Suppose this claim is indeed correct. Yet, when we randomly sampled 100 individuals, we obtained a test statistic that led us to reject the claim. What should we conclude? We should REJECT the null hypothesis. We should FAIL TO REJECT the null hypothesis. We have committed a Type I Error. We have committed a Type II Error. Since our test statistic lies in the rejection region, we should repeat the experiment until we can consistently FAIL TO REJECT. 4 _____ 26. Suppose a researcher’s claim is false yet we fail to reject their claim. What type of error is this? Type I Error Type II Error Type III Error _____ 27. Suppose we seek to estimate the population mean price for mobile homes by obtaining a simple random sample of 36 homes and recording the price. Further, we find that the sample mean price of these 36 homes is 59.28 (measured in thousands). What does 59.28 represent? Sample mean Statistic Point Estimate Only parts and All of the above. . _____ 28. A college statistics professor has office hours from 9:00 A.M. to 10:30 A.M. daily. A sample of thirty waiting times to see the professor (in minutes) is 10, 12, 20, 15, 17, 10, 30, 28, 35, 28, 19, 27, 25, 22, 33, 37, 14, 21, 20, 23, 22, 25, 27, 28, 29, 29, 29, 30, 31, 32 Assuming 7.84, what is the 95.44% confidence interval for the population mean? 22.39 to 27.08 minutes 21.46 to 27.07 minutes 18.8 to 25.8 minutes 16.84 to 29.32 minutes _____ 29. What is the value of that corresponds to a confidence level of 85%? 15 0.15 0.015 0.85 _____ 30. Ten randomly selected students took a final exam in psychology in a large class. If the sample mean was 78 and the sample standard deviation was 8.0, what is the 99% confidence interval for the mean score of all students? Assume the population of scores for the class is normally distributed. 69.78 to 86.22 74.40 to 81.60 73.97 to 82.03 73.99 to 82.01 31. Suppose we wish to estimate the mean household income in Arlington. To carry out an analysis, we randomly select 100 households in the area and obtain a sample mean income of $78,000. We then conclude that the mean household income in the Arlington area is exactly $78,000 (with sample standard deviation $4,250). Is this problematic? If yes, do you suggest an alternative analysis? 32. What is the implied meaning of the expression “statistical significance?” Note: This is technical terminology in the study of statistics. 5 Exam 3 Review: Solutions 1. Use a Z‐Test since the population standard deviation is known. This is a two‐tailed test since the alternative hypothesis is NOT equal to the null hypothesis (thus we’ll reject the claim if our sample mean is “too low” or “too high”). Critical Values: InvNorm(.025) = 1.96 AND ‐1.96. Note: We use “.025” since this is a two‐tailed test and .05/2 = .025! Test Statistic is given by Since ‐2.40 < ‐1.96, we REJECT the null hypothesis (i.e. the report’s claim). Thus, at the 5% level of significance, we reject the claim that the mean household income in Fairfax County is equal to $105,000. In other words, we have enough evidence to conclude that the mean household income in Fairfax County is NOT $105,000! ‐2.40 ‐1.96 0 +1.96 T.S. C.V. Report’s Claim C.V. Please note the following: •
This is a sketch of SAMPLE MEANS! Hence we are comparing sample means (of size 36). •
Since the mean of these sample means equals the population mean, and the report claims the population mean is $105,000 (with a corresponding z‐score of zero), we mark the claim at the center of the graph (always!). •
The test statistic corresponds to your obtained sample mean of size 36 households. •
The critical value corresponds to the “cut‐off” score. This value separates the rejection region from the non‐rejection region. The p‐value approach compares a probability with the level of significance. To find this value: Normalcdf(‐100000,‐2.4) = .008 Since this is a two‐tailed test, multiply this number by 2 to get .0164. Assuming that the null hypothesis is true, this tells us that the probability of observing a sample mean of $95,000 or less (or $115,000 or more, since $115,000 has a z‐score of +2.4) is .0164. THIS IS UNLIKELY! Since .0164 < .05, we REJECT the claim (null hypothesis). Please note the following: •
[Calculator] You may also use the STAT > TESTS > [1] Z‐Test to obtain the p‐value and test‐statistic. •
[Z‐Table] You may also use a z‐table to obtain the p‐values (look for the area to the left of ‐2.40 and multiply by 2). 6 Area to left of test statistic is .008 ‐2.40 0 2.40 Area to right of test statistic is .008 Again, with the report’s claim being $105,000, this tells us that the probability of obtaining a sample mean less than or equal to $95,000 (z‐score less than or equal to ‐2.40) OR greater than or equal to $115,000 (z‐score greater than or equal to 2.40) is .008 + .008 = .0164. Since the level of significance is .05 (we’ll reject their claim if our sample mean falls in either tail with 2.5%), and .0164 < .05, we REJECT! Critical Values: invNorm(.005) = 2.576 AND ‐2.576; Test‐statistic = ‐2.40. •
Using the critical value approach, since ‐2.40 > ‐2.576, we FAIL TO REJECT the claim. •
Using the p‐value approach, since .0164 > .01, we FAIL TO REJECT the claim. That is, at the 1% level of significance, there is NOT ENOUGH evidence to reject the claim that the mean household income is $105,000. 2. Use a T‐Test since a sample standard deviation is given (and population standard deviation is not known). This is a one‐tailed test (right‐tail) since the alternative hypothesis is GREATER than the proposed mean (thus we’ll reject the claim if our sample mean is “significantly higher” than 0.5). Using a t‐table, look up df = 16 – 1 = 15 with area .05 (Be careful: do NOT divide by 2!). The critical value is 1.753. Test Statistic is given by Since 2.0 > 1.753, we REJECT the null hypothesis (i.e. the department’s claim). Thus, at the 5% level of significance, we reject the claim that the mean consumption of mean is equal to 0.5 lbs. In other words, we have enough evidence to conclude that the mean meat consumption of a given American is MORE than 0.5 lbs (our intuitions were correct!). Department’s Claim C.V. T.S. 0 7 +1.75 2.0 Please note the following: •
This is a sketch of SAMPLE MEANS! THIS IS A T‐DISTRIBUTION (APPROXIMATELY NORMAL)! •
Since the mean of these sample means equals the population mean, and the report claims the population mean is 0.5 lbs. (with a corresponding z‐score of zero), we mark the claim at the center of the graph (always!). •
The test statistic corresponds to your obtained sample mean of size 16 individuals. •
The critical value corresponds to the “cut‐off” score. This value separates the rejection region from the non‐rejection region. The p‐value approach compares a probability with the level of significance. To find this value (you CANNOT use Normalcdf!), look in your t‐table across df = 15 and try to locate the t‐statistic 2.0. Notice that we only have 2.132 and 1.753. We could try to average these two values to get a rough estimate for 2.0, but this value may be a bit inaccurate. Instead, we resort to using our calculator: STAT > TESTS > [2] T‐TEST 0.50; .55; s = .1; n = 16; highlight “ ” We obtain a p‐value of .0320. Since .0320 < .05, we REJECT the null hypothesis in favor of the alternative hypothesis. Intuitively, this implies it is improbable to obtain a sample mean of 0.55 lbs assuming the population mean is 0.50 lbs. Critical Value: 2.602; Test‐statistic: 2.0 •
Using the critical value approach, since 2.602 > 2.0, we FAIL TO REJECT the claim. •
Using the p‐value approach, since .0320 > .01, we FAIL TO REJECT the claim. That is, at the 1% level of significance, there is NOT ENOUGH evidence to reject the claim that the mean meat consumption is half pound. 7. 156.2;
5.0625 8. [Central Limit Theorem Part I] Since the population is normally distributed, then the distribution of sample means (each of size 29 male brains) will also be normal. 9. [Central Limit Theorem Part II] Since the sample size is greater than or equal to 30, then the distribution of sample mean salaries (of sample size 45 players each) will be normal! It not necessary that the population be normal. 10. Since the population is normally distributed, the distribution of sample means will also be normal by the Central Limit Theorem [Part I]. The probability that the sample mean rent is less than $1800 is .9332. [Calculator: normalcdf(‐1000, 1800, 1500, 200) = .9332] 11. Student .4602 Calculator: normalcdf ‐10000, 44, 45, 10 .4602 ; Sample Mean .2420 Calculator: normalcdf ‐10000, 44, 45, 10/√49 .2420 12. 2.131 13. n – 1 15 – 1 14 14. ‐10.8 15. Z‐Test; the population standard deviation is known; simple random sampling is utilized; large sample (and thus we don’t need a normal population). 16. Neither; sample standard deviation is known; simple random sampling; but if outliers are present, we need the population to be normal OR a large sample, but we have neither here. 17. Neither; we MUST have simple random sampling! If we had this, then a Z‐Test would be appropriate. 18. To test the validity of a claim (made by an individual, an advertisement, a company, previous research, etc.) by looking at a sample. 19. [Critical‐Value Approach] A value that separates the r...
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 Fall '10
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 Math, Standard Deviation, null hypothesis

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