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Assignment 3 week 9-week 11 Student Full Name:___________________________________ . Student ID:__________________________________________ . CRN No:____________________________________________ . Branch: _____________________________________________. STATISTICS (STAT-101) Total Points True/False ____/6 MCQ ____/6 Short Answer ____/18 Total ____/30 Good Luck STATISTICS (STAT-101) Marks- 30 Answer all the Questions on the same question paper. Section-I State whether the following statements are True or False. (6 marks, 1 Mark Each) 1. A type I error is the mistake of rejecting the null hypothesis when it is actually true. True 2. If P βvalue is greater than Ξ± (level of significance) reject the null hypothesis. False 3. Two samples are dependent if the sample values are matched pairs. True 4. In case of hypothesis testing for a sample, the t statistic is used if π is not known and sample size n is greater than 30 (n > 30). True 5. A claim that two population proportions are equal, each of the two samples must satisfy the requirement that ππ β₯ 5 and ππ β₯ 5. True 6. In an unpaired samples t-test with sample sizes π1 = 21 and π2 = 11, the value of t should be obtained at 10 degree of freedom. True Section-II (Multiple Choice Questions) (6 marks, 1 Mark Each) 1. Area of the rejection region depends on β¦ a. Size of Ξ± b. Size of Ξ² c. Test-Statistic. d. Number of values 2. A two-tailed test is one where: a. results in only one direction can lead to rejection of the null hypothesis b. negative sample means lead to rejection of the null hypothesis c. results in either of two directions can lead to rejection of the null hypothesis d. no results lead to the rejection of the null hypothesis Page 2 of 10 3. A decision in a hypothesis test can be made by using a a. P-value b. Critical Value c. A and B d. None of the above 4. When carrying out a large sample test of H0: π0 = 10 vs. Ha: π0 > 10 by using a rejection point, we reject H0 at level of significance πΌ when the calculated test statistic is: a. Less than π§πΌ b. Less than - π§πΌ c. Greater than π§πΌ/2 d. Greater than π§πΌ 5. A randomly selected sample of 500 college students was asked whether they had ever used the drug Ecstasy. Sixteen percent (16% or 0.16) of the 500 students surveyed said they had. Which one of the following statements about the number 0.16 is correct? a. It is a population proportion. b. It is a margin of error. c. It is a sample proportion. d. It is a randomly chosen number. 6. Two teams of workers assemble automobile engines at a manufacturing plant in Michigan. A random sample of 145 assemblies from team 1 shows 130 acceptable assemblies. A similar random sample of 125 assemblies from team 2 shows 120 acceptable assemblies. The pooled proportion for acceptable assemblies is: 250 a. b. c. d. 270 270 250 20 270 270 20 Part-II (Multiple Choice Questions) MCQ 1 2 3 4 Answers A C C D (6 marks, 1 Mark Each) 5 C Page 3 of 10 6 A Section βIII Answer the following Essay Type Questions (18 marks, 3 Mark Each) 1. A medical researcher claims that less than 20% of American adults are allergic to a medication. In a random sample of 100 adults, 15% say they have such an allergy. Test the researcherβs claim at ο‘ = 0.01. Solution: The products ππ = 100 × 0.20 = 20 and ππ = 100 × 0.80 = 80 are both greater than 5. So, we can use the z-test. The claim is βless than 20% are allergic to a medication.β Step 1 The claim is π < 0.2 Step 2 Alternative to claim is π β₯ 0.2 Step 3 the null and alternative hypothesis are: π»0 : π = 0.2. π»π : π < 0.2 (Claim) Step 4 the level of significance is ο‘ = 0.01 Step 5 Using the z-test, the standardized test statistic is: zο½ pΛ ο p ο½ pq n 0.15 ο 0.20 ο» ο1.25 (0.20)( 0.80) 100 Step 6: Because the test is a left-tailed test and level of significance is ο‘ = 0.01, the critical value is π0 = β2.33 and the rejection region is the area where π < β2.33. Page 4 of 10 The graph shows the location of the rejection region and the standardized test statistic. Because z is not in the rejection region, you should decide fail to reject the null hypothesis. Or P-value is the area to the left of z = -1.25; The P-value of 0.1056 is greater than the significance level of ο‘ = 0.01, fail to reject the null Hypothesis. In other words, there is not enough evidence to support the claim that less than 20% of Americans are allergic to the medication. 2. Suppose we would like to determine if the typical amount spent per customer for dinner at a new restaurant in town is more than $20.00. A sample of 49 customers over a three-week period was randomly selected and the average amount spent was $22.60. Assume that the standard deviation ο³ is known to be $5.50. Using a 0.05 level of significance, would we conclude the typical amount spent per customer is more than $20.00? Solution: ο³ is known ( $5.50), sample size is 49 (n > 30) Step 1 The claim is π > 20 Step 2 Alternative to claim is π β€ 20 Step 3 The null and alternative hypothesis are: π»π: π = 20, π»π: π > 20 Step 4 the significance level is ο‘ = 0.05 Page 5 of 10 Step 5 ο³ = 5.5, π = 49 π₯Μ = 22.60, π = 20 Zο½ xοο ο³ n ο½ 22.60 ο 20 ο½ 3.30 5.50 49 Step 6: Right-tailed test, so P-value is the area is to the right of z = 3.30; The P-value of 0.0005 is less than the significance level of ο‘ = 0.05, reject the null Hypothesis. P-value = 0.0005 ο = 20 or z=0 z = 3.30 There is sufficient evidence to conclude the typical amount spent per customer is more than $20.00. 3. The scores on an aptitude test required for entry into a certain job position have a mean at most 500. If a random sample of 36 applicants have a mean of 546 and a standard deviation of 120, is there evidence that their mean score is different from the mean that is expected from all applicants?. Use a 0.05 level of significance. Solution: population standard deviation is not known, sample size is 36 (n > 30) Page 6 of 10 Step 1 The claim is π β€ 500 Step 2 Alternative to claim is π > 500 Step 3 Null and Alternative Hypothesis are: π»0 : π = 500 π»π : π > 500 . Step 4 the significance level is ο‘ = 0.05 Step 5 π = 120, π = 36 π₯Μ = 546, π = 500 tο½ x ο ο 546 ο 500 ο½ ο½ 2.3 s 120 n 36 Step 6 The degrees of freedom (df=n-1) is 36-1=35. Using t-table with 35 degrees of freedom and level of significance 0.05, the t-value is 1.690. t = 2.3 fall in the critical region bounded by t = 1.690. Reject the null hypothesis. ο = 500 or z=0 t = 2.3 Critical value t = 1.690 Page 7 of 10 There is sufficient evidence to conclude that the mean score is different from the mean that is expected from all applicants. 4. Suppose a researcher believes that college faculty vote at a lower rate than college students. She collects data from 200 college faculty and 200 college students using simple random sampling. If 120 of the faculty and 150 of the students voted in the 2016 Presidential election, is there enough evidence at the 5% level of significance to support the researcherβs claim? Solution: two samples are independent and π₯1 = 120, π1 = 200. π₯2 = 150, π2 = 200. Consider p1: proportion of successes for population 1(Faculty) p2: proportion of successes for population 2(Students) Step 1 The claim is π1 < π2 Step 2 Alternative to claim is π1 β₯ π2 Step 3 The null and alternative hypothesis are: π»0 : π1 = π2 . π»π : π1 < π2 (Claim) Step 4 significance level is ο‘ = 0.05 Step 5 π₯1 = 120, π1 = 200. π₯2 = 150, π2 = 200. πΜ = zο½ π₯1 +π₯2 π1 +π2 = 0.675, πΜ = 0.325 ο¨pΜ1 ο pΜ2 ο©ο ο¨p1 ο p2 ο© pq pq ο« n1 n2 120 150 β 200 200 π§= = β3.2 0.675 × 0.325 0.675 × 0.325 β + 200 200 Page 8 of 10 So the P- value is 0.0007. Step 6 The P-value of 0.0007 is less than the significance level of ο‘ = 0.05, we reject the null hypothesis of π1 = π2 . Because we reject the null hypothesis, we conclude that there is sufficient evidence to support the claim. Q. 5 & 6.Use the following information to answer Questions 5 and 6: Given the following data of two independent samples of normally distributed populations Data Population 1 Population 2 n 23 13 43 41 4.5 5.1 π₯Μ s 5. Test the claim that π1 β π2 at the Ξ±=0.05 level of significance 6. Construct a 95% confidence interval about π1 β π2 . Solution(5) : two samples are independent and π₯1 = 43, π1 = 23, π 1 = 4.5. Μ Μ Μ π₯2 = 41, π2 = 13, π 2 = 5.1. Μ Μ Μ Consider π1 : mean of population 1 π2 : mean of population 2 Step 1 The claim is π1 β π2 Step 2 Alternative to claim is π1 = π2 Step 3 The null and alternative hypothesis are: π»0 : π1 = π2 . π»π : π1 β π2 (Claim) Step 4 Significance level is ο‘ = 0.05 Page 9 of 10 Step 5 π₯1 = 43, π1 = 23, π 1 = 4.5. Μ Μ Μ π₯2 = 41, π2 = 13, π 2 = 5.1. Μ Μ Μ π‘= (π₯ Μ Μ Μ 1 β Μ Μ Μ ) π₯2 β (π1 β π2 ) π 2 π 2 β 1+ 2 π1 π2 π‘= 43 β 41 β4.5 × 4.5 + 5.1 × 5.1 23 13 = 1.178 Step 6: The t- value at πΌ/2 = 0.025 and df=n-1=13-1=12 is ±2.179. Because the test statistic does not fall within the critical region, fail to reject the null hypothesis: π1 = π2 . Solution(6): Margin of error (E) is given by πΈ= π 12 β πΌ π‘ β2 π1 π 22 + π2 Thus 4.5 × 4.5 5.1 × 5.1 πΈ = 2.179β + = 2.179 × 1.697 = 3.699 23 13 Confidence interval is given by (π₯ Μ Μ Μ 1 β Μ Μ Μ ) π₯2 β πΈ < (π1 β π2 ) < (π₯ Μ Μ Μ 1 β Μ Μ Μ ) π₯2 + πΈ (43 β 41) β 3.699 < (π1 β π2 ) < (43 β 41) + 3.699 β1.699 < (π1 β π2 ) < 5.699 Page 10 of 10