ALL-ANSWERS (1)

ECON 2330 SAMPLE MIDTERM QUESTIONS Multiple Choice: 1. The producer of a certain bottling equipment claims that the variance of all their filled bottles is 0.027 or less. A sample of 30 bottles showed a standard deviation of 0.2. The p -value for the test is a) between 0.025 to 0.05 b) between 0.05 to 0.01 c) 0.05 d) 0.025 2. What type of error occurs if you fail to reject H o when, in fact, it is not true? a) either Type I or Type II, depending on the level of significance b) Type II c) either Type I or Type II, depending on whether the test is one-tailed or two-tailed d) Type I 3. A sample of 41 observations yielded a sample standard deviation of 5. If we want to test H o : ! 2 = 20, the test statistic is a) 100 b) 10 c) 51.25 d) 50 4. Salary information regarding male and female employees of a large company is shown below. Male Female Sample Size 64 36 Sample Mean Salary (in $1,000) 44 41 Population Variance 128 72 The point estimate of the difference between the means of the two populations is a) -28 b) 3 c) 4 d) -4 5. A sample of 51 elements is selected to estimate a 95% confidence interval for the variance of the population. The chi-square values to be used for this interval estimation are a) -1.96 and 1.96 b) 32.357 and 71.420 c) 34.764 and 67.505 d) 12.8786 and 46.9630 6. If the cost of making a Type I error is high, a smaller value should be chosen for the a) confidence coefficient. b) test statistic. c) level of significance. d) critical value.

2 7. We are interested in testing whether the variance of a population is significantly less than 1.44. The null hypothesis for this test is a) H 0 : σ 2 < 1.44 b) H 0 : s 2 ≥ 1.44 c) H 0 : σ < 1.20 d) H 0 : σ 2 ≥ 1.44 8. A statistics teacher wants to see if there is any difference in the abilities of students enrolled in statistics today and those enrolled five years ago. A sample of final examination scores from students enrolled today and from students enrolled five years ago was taken. You are given the following information Today Five years ago "̅ 82 88 ! ! 112.5 54 n 45 36 The 98% confidence interval for the difference between the two-population means is a) -13.84 to -1.16 b) -10.66 to -1.34 c) -9.92 to -2.08 d) -24.77 to 12.23 9. The value of F .05 with 8 numerator and 19 denominator degrees of freedom is a) 2.48 b) 2.58 c) 3.63 d) 2.96 10. If the null hypothesis is rejected at the .05 level of significance, it will ____ be rejected at the .10 level of significance. a) always b) sometimes c) always not d) sometimes not 11. Generally, the ____ sample procedure for inferences about two population means provides less noise than the ____ sample approach. a) Single, independent b) Matched, independent c) Matched, pooled d) Independent, pooled 12. The producer of a certain medicine claims that their bottling equipment is very accurate and that the standard deviation of all their filled bottles is .2 ounces or less. A sample of 20 bottles showed a standard deviation of .12 ounces. The test statistic to test the claim is a) 2.3 b) 22.99 c) 6.84 d) 1.368

4 19. Sample A Sample B s 2 40 96 n 16 26 We want to test the hypothesis that the population variances are equal. The test statistic for this problem equals a) 0.417 b) 0.843 c) 2.4 d) 1.500 20. A convenience store has an average sale of $8000 per day. The store introduced several promotional campaigns to boost sales. To determine whether or not the campaigns have been effective in increasing sales, a sample of 100 days was selected, and it was found that the average was $8200 per day. Suppose we know ! = $1500. The p-value is a) 0.1333 b) 0.0918 c) 0.9082 d) 1.3333 21. The sampling distribution of the ratio of independent sample variances extracted from two normal populations with equal variances is the a) chi-square distribution b) normal distribution c) F distribution d) t distribution 22. The following information was obtained from matched samples Individual Method 1 Method 2 1 7 5 2 5 9 3 6 8 4 7 7 5 5 6 The point estimate for the difference between the means of the two populations is (Method 1- Method 2) a) -1 b) 0 c) 1 d) 2 23. The results of a recent poll on the preference of shoppers regarding two products are shown below. Product Shoppers Surveyed Shoppers Favoring this product A 800 560 B 900 612 At 95% confidence, the margin of error is a) 0.064 b) 0.0225 c) 0.025 d) 0.044

5 24. The p-value a) Can be any negative value b) Can be any positive value c) Must be a number between 1 and 0 d) Must be a number between -1 and 0 25. A Type II error is committed when a) a true null hypothesis is mistakenly rejected b) the true null hypothesis is correctly rejected c) a true alternative hypothesis is mistakenly rejected d) the true alternative hypothesis is correctly rejected 26. To avoid the problem of not having access to tables of the F distribution when a one- tailed test is required and with F values given for the lower tail, let the a) smaller sample variance be the numerator of the test statistic b) larger sample variance be the numerator of the test statistic c) sample variance from the population with the smaller hypothesized variance be the numerator of the test statistic d) sample variance from the population with the larger hypothesized variance be the numerator of the test statistic 27. If we are interested in testing whether the proportion of items in population 1 is larger than the proportion of items in population 2, the a) null hypothesis should state P1 - P2 < 0 b) null hypothesis should state P1 - P2 ³ 0 c) alternative hypothesis should state P1 - P2 > 0 d) alternative hypothesis should state P1 - P2 < 0 28. A random sample of 25 students selected from a university had an average age of 25 years and a standard deviation of 2 years. We want to determine if the average age of all the students at the university is significantly more than 24. Assume the distribution of the population of ages is normal. The test statistic is a) 0.05 b) 1.38 c) 2.50 d) 1.56 Questions: 1. Most individuals are aware of the fact that the average annual repair cost for an automobile depends on the age of the automobile. A researcher is interested in finding out whether the variance of the annual repair costs also increases with the age of the automobile. A sample of 26 automobiles 4 years old showed a sample standard deviation for annual repair costs of $170 and a sample of 25 automobiles 2 years old showed a sample standard deviation for annual repair costs of $100. a) State null and alternative hypotheses that the variance in annual repair costs is larger for older automobiles. b) At a 0.10 level of significance, what is your conclusion?

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Q1) a) 𝑞 ̂𝑑 = 908.47 + 82.56𝐺 − 8.9𝑃? − 10.35𝑃? + 7.41𝐴 b) Price of the good, price of a related good, advertising expenditure and Gender together capture 72% of the variation in Quantity demanded. c) Interpreting co-efficient For every $1 increase in price and price of a related good, the average Quantity demanded decreases by 8.9 units and 10.35 units respectively. For every $1000 increase in Advertising Expenditure, the average Quantity demanded increases by 7.41 units. Quantity demanded from Women is on average 85.56 units more than men. Interpreting corresponding confidence intervals Gender: I am 95% confident that Qd from women is on average between 17.99 and 147.13 more than men. Px: I am 95% confident that for every $1 increase in Px, the average Qd will fall between 12.28 and 5.53 units Py: I am 95% confident that for every $1 increase in Py, the average Qd will fall between 14.73 and 5.97 units A: I am 95% confident that for every $1000 increase in A, the average Qd will increase between 4.06 and 10.77 units.

Q2) Ho: 𝜇 𝑅 = 𝜇 𝑀 = 𝜇 𝐺 Ha: Not all are equal/ at least one is different 𝛼 = 0.1 Sources of Variation SS df MS F P-vlaue Treatments 6790 2 (k-1) 3395 3.61 0.05 < p < 0.1 Blocks 323038.37 4 (b-1) 80759.59 Error 7531.63 8 (k-1)(b-1) 941.45 Total 337360 14 (n-1) i) SSTR = 5[ (714 − 741) 2 + (743 − 741) 2 + (766 − 741) 2 ] = 6790 ii) SSBL = 3[ (871.67 − 741) 2 + (758.33 − 741) 2 + (800 − 741) 2 + (456.67 − 741) 2 + (818.33 − 741) 2 ] = 323038.37 iii) SSE = SST – SSTR – SSBL = 7531.63 iv) MSTR = SSTR/df v) MSBL = SSBL/df vi) MSE = SSE/df vii) F= MSTR/MSE Since p< 0.1 Reject Null FISHERS LSD Test Test statistic LSD = ( 𝑡 0.05 8 ) √[𝑀𝑆𝐸 ( 1 𝑛 ? + 1 𝑛 ? )] Ho: 𝜇 𝑅 = 𝜇 𝑀 Ha: 𝜇 𝑅 ≠ 𝜇 𝑀 714-743 = 29 < DNR 36.09 Ho: 𝜇 𝑅 = 𝜇 𝐺 Ha: 𝜇 𝑅 ≠ 𝜇 𝐺 714-766 = 52 > REJECT 36.09 Ho: 𝜇 𝑀 = 𝜇 𝐺 Ha: 𝜇 𝑀 ≠ 𝜇 𝐺 743 – 766 = 23 > DNR 36.09 * 𝑡 0.05 8 = 1.860 Average grades in reading and math are similar and average grade in Grammar is different.