GE6

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Apr 3, 2024

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STAT 200 Guided Exercise 6 For On-Line Students, be sure to: • Submit your answers in a Word file at the same place you downloaded the file • Remember you can paste any Excel or JMP output into a Word File (use Paste Special for best results). • Put your name and the Assignment # on the file name: e.g. Ilvento Guided6.doc Key Topics • Hypothesis Tests for Single Means • Hypothesis Tests for Single Proportion Answer as completely as you can and show your work. Then upload the file to get credit. 1. The EPA standard on the amount of suspended solids that can be discharged into a river is a maximum of 60mg per liter per day (mg/L). You want to test a randomly selected sample of n water specimens and estimate the mean daily rate of pollution produced by a mining operation. You have been asked to submit a proposal to do this study. The granting agency wants to know about the precision of your estimate Suppose you want a 95% C.I. with a bound of error (B) of .8 mg/l. The formula for solving for n is to the right. Please note that it is an algebraic manipulation of the confidence interval formula where we solve for n, based on a desired plus or minus bound of error of .8 mg/l. The mean of the sample doesn’t actually play any role here. I do need a good estimate of Sigma to execute this formula. We will use σ = 5 mg/L . a. Solve the equation for n. What is the minimum sample size you would need? Assume the water readings are approximately normal with σ = 5 mg/L . Always round up for these problems. We start with the following: z= 1.96 since α= .05; B = .8 mg/l; σ = 5 mg/L n = ( ? . ??? ) ?? . ?? = ( ? . ???? ) ?? . ?? = n = 150.06 n = 151 as a minimum sample size needed (we round up to be safe). b. Check you answer by calculating the confidence interval for this problem (the Bound of Error) using the sample size you calculated. ± ? . ?? ( ? / ??? ) = ± . ???? ≈ ± . ? We show that a sample size of 151 will give a BOE of .8 (or less) on a 95% C.I n = ( z a / 2 ) 2 s 2 B 2

2. A larger supermarket began to get complaints from customers about the quantity of chips in a 16-ounce bag of a particular brand of potato chips. The chain decided to test the following hypothesis concerning  , the mean weight (in ounces) of a bag of potato chips in the next shipment of chips received from their supplier. H o :  = 16 H a :  < 16 If there is evidence that  < 16, then the shipment would be refused and a complaint would be registered with the supplier. a. What is the Type I error, expressed in terms of this problem. Say it in words. Refusing the shipment when the weight of the chips was ok. b. What is Type II error, expressed in terms of this problem. Accepting the shipment when in fact the chips were underweight c. Which type of error would the customers view as more serious? The customers would rather err on the side of being cautious and reject more shipments- They would be concerned with Type II Error. d. Which type of error would the supplier view as more serious? The supplier would rather have their product accepted – they would be concerned with Typw 1 Error, rejecting shipments when the weight was ok 3. A car manufacturer wants to test a new engine to determine if it meets air pollution standards. The mean emissions of all engines of this type must be less than 20 parts per million of carbon. Ten engines are manufactured for testing purposes and the emissions are measured. We will assume that this distribution is approximately normal. This is a small sample hypothesis test of a mean. Carbon 15.6 Sum(X) = 175.7 12 | 7 16.2 Sum(X 2 ) = 3165.49 13 | 22.5 N = 10 14 | 9 20.5 15 | 6 16.4 16 | 2 4 19.4 17 | 9 19.6 18 | 17.9 19 | 4 6 12.7 20 | 5 14.9 21 | 22 | 5

a. Calculate a 95% confidence interval for this sample estimate. Standard Error = 2.952/ (10).5 = .934; t.05/2, 9 d.f. = 2.262. Since n is small, the t-value is much larger than what we would have used with a z-value (1.96). 17.570 ± 2.262(.934) 17.570 ± 2.112 15.458 to 19.682 b. Test to see if the sample data is lower than the emissions requirement of 20 parts per million. You will need to determine the Null Hypothesis and your Alternative Hypothesis. Use α = .05. Note: use the t-distribution for the rejection region. HYPOTHESIS TEST Null Hypothesis H0 : μ = 20 Alternative Hypothesis Ha : μ < 20 One -tailed Test, Lower Why, one-tailed? Test to see if the sample data is lower than the Assumptions of Test n = 10 this a small sample. I use the t- distribution, with α= 0.05 and 10-1 = 9 d.f. We assume the distribution in the population is approximately normal. Test Statistic (z* or t*) t * = (17.570 – 20)/.93 Rejection Region t.05,9 d.f. = -1.833 Compare this to z-value of 1.645 Calculation of Test Statistic t* = -2.603 Comparison of Test Statistics with Rejection Region t * < t.05,9 d.f. -2.603 < -1.833 Our Test Statistic, t*, is in the Rejection Region. t* is negative, and below the Critical Value set for the left-hand tail. As a result, we have evidence, within our accepted probability framework, to Reject Ho: μ = 20. Excel and JMP gives a p-value of .0143 for this one-tailed test. This p- value is less than α = .05 for the test, so we also Reject Ho: μ = 20. The conclusion from the Critical Value/Rejection Region approach will always agree with the p-value approach. They are part and parcel of the same test.

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c. We will change two components of this test to see how the result would change. • What if we used a two-tailed Alternative Hypothesis for this test. Would the conclusion of the test change? Ha : μ ≠ 20 Two -tailed Test • the Critical Values are -2.262 and 2.262. Note: these are the same values as for the Confidence Interval for the same level of alpha. • Since t* = - 2.603, the conclusion of the test remains the same, Reject Ho: μ = 20. • The p-value for the test becomes .0286 • What if they wanted alpha to be .01 instead of .05. Keep the test as a one tailed test, but change the level of alpha. Would the conclusion of the test change? Everything stays the same, except: • The level of alpha. It changes from .05 to .01 • The Critical Value for the test. It changes from -1.833 to -2.821 • Possibly the outcome of our test. With this shift: Hypothesis Test using t-distribution Sample Mean 17.570 Sample std dev 2.952 Sample Size 10.00 Null Value 20.000 Alpha Value 0.050 d.f. 9.00 Std Error 0.934 T.S. t* -2.603 Conclusion p-value 1-tailed 0.0143 Reject Ho: p-value 2-tailed 0.0286 Reject Ho: Lower Upper 1-tailed Critical t Value -1.833 1.833 2-tailed Critical t Value -2.262 2.262  t* > t.01,9 d.f. -2.603 > -2.821  Our Test Statistic, t*, is not in the Rejection Region. t* is negative, and above the Critical Value set for the left-hand tail. As a result, we DO NOT have evidence, within our accepted probability framework, to Reject Ho: μ = 20. We fail to reject

4. A large city’s Department of Motor Vehicles (DMV) claimed that 80% of candidates pass driving tests on the first try . A newspaper surveyed 90 randomly selected local teens who had taken the test and found 61 teens had passed on the first try. This value seemed lower than the one reported by DMV. Test to see if the passing rate for teenagers from our sample is less than the overall rate. This is a Hypothesis Test for a Proportion. a. What is the proportion for our sample? Estimated p = 61/90 = .6778 b. Determine if we can use the normal approximation of the binomial. Remember, n*p or n*q should be > 5 in order to use this approach. 90*.6778 = 61 90*.3222 = 29. Yes we can use the normal approximation to the Binomial for this data c. If the Null Hypothesis is .80, what is the standard error for this problem? Use the value under Ho: p = .80. S.E. = SQRT[(.80*.20)/90] = SQRT[.0018] = .0422 d. Now finish the steps for the hypothesis test. Use an alpha level of .05. HYPOTHESIS TEST Null Hypothesis Ho: P = .80 Alternative Hypothesis Ha: P < .80. One-tailed Lower Remember, it said: Test to see if the passing rate for teenagers from our Assumptions of Test Sample size is large enough, use the Normal Approximation to Binomial. Test Statistic (z* or t*) z* = (.6778 - .80)/.0422 Rejection Region z.05 = -1.645 Calculation of Test Statistic z* = -2.899 Comparison of Test Statistics with Rejection Region z* < z.05 ; -2.899 < -1.645. Reject H0: p = .80 ; We have evidence that the sample proportion is lower than .80. e. Take the test statistic that you calculated (z*) and find the probability associated with this test statistic value in the standard normal table. The probability of the test statistic is thought of the probability of that value and further out in the tail. For this problem, it is .5 - p(ABS[z*]). We use the absolute value since our z-score is negative. This value is known as a p-value for this problem. • We want to compare the p-value to the alpha level set for the problem. If it is lower than alpha, we can reject the Null Hypothesis. Does the p-value approach match your conclusion above? With a z* = -2.90 the p-value is: .5 - .4981 = .0019 This is much lower than a = .05, so we can reject the Null Hypothesis

• What if our alternative hypothesis was a two-tailed test, what would the p-value be? f this were a two-tailed test, we would double the p-value for a one-tailed test. p= .0019*2 = .0038. This is still lower than .05. We would still reject the Null Hypothesis. 5. Therapeutic touch, taught in many schools of nursing, is a therapy in which the practitioner moves her hand near, but does not touch, a patient in an attempt to manipulate a “human energy field.” Therapeutic touch practitioners believe that by adjusting this field they can promote healing. A study was done to see if there is any evidence of Therapeutic touch (TT). The experiment involved 15 TT practitioners who were asked to determine if they could detect an unseen hand hovering over their right or left hand. Practice sessions were used at first where the practitioners could see the hovering hand. Then a screen was placed over the hands and 10 trials were conducted for each practitioner. A coin flip was used to randomly determine which hand to hover over. Overall, 150 trials were conducted and the TT practitioners were successful 70 times, for a proportion of .467. a. Is there evidence from this experiment that TT practitioners can successfully detect a “human energy field?” You need to conduct a hypothesis test for this proportion problem. Use an alpha level of .01 . Start with what you know. • A chance guess should result in P = .5. That should be our Null Hypothesis. • Based on the Null Hypothesis, the Standard Error for this problem would be: o S.E. = SQRT(.5*.5)/150) = .0408 • Our sample estimate is p = 70/150 = .4667. Notice, we expected it to be greated than .5 and our estimate is smaller than .5 • Logically, we would expect this to be a one-tailed upper test. TT Practitioners should do better than chance HYPOTHESIS TEST Null Hypothesis Ho: P = .5 Alternative Hypothesis Ha: P > .5 Assumptions of Test Sample size is large enough, use the Normal Approximation to Binomial Test Statistic (z* or t*) z* = (.4667 - .50)/.0408 Rejection Region z.01 = 2.33

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Calculation of Test Statistic z* = -.8165. Notice, we expected it to be positive and it is negative Comparison of Test Statistics with Rejection Region z* < z.01 -8165 < 2.33 Cannot Reject Ho: p = .50 We do not have evidence that the sample proportion is larger than .50. b. Calculate the p-value for this test. So the p-value is bases on z* = -.8165. If we look this up in the standard normal table we have a probability of Lookup in table .2939. But we need to add the .5 for the right-hand side of the distribution. So the final p-value is quite large at .7939. That is what happens when you anticipate a one-tailed test before collecting your data only to find that your estimate is in the other side of the distribution. 6. Speeders on the highway. The following is data from a random sample of 63 motorists on a Federal highway where the speed limit of 65 mph. The data can be found on an Excel file on the web (speeders.xls). a. Below is the JMP output from the descriptive statistics of the speed of drivers. Briefly (one paragraph) describe the data using the summary statistics. The mean speed for the 63 motorists was 69.22 mph with a range of 39. The highest speed was 95 and the lowest was 56. The distribution is approximately mound shaped and symmetrical with a slight right skew. The mean, median, and mode are close to each other (69.22, 67, and 65), however the mean is slightly higher and is being pulled by a few extreme high values. The standard deviation for this data is 8.07 and the CV is 11.65. There is not a large amount of variability.

b. Conduct a Hypothesis Test where the null hypothesis is that the average speed is 65 mph. Let’s let this be a two- tailed test. Use an α level of .01. Everything you need to do the test is given in the JMP output. HYPOTHESIS TEST Null Hypothesis μ = 65 Alternative Hypothesis μ ≠ 65 Two -Tailed Tes Assumptions of Test The sample size is relatively large (63) so we do not need to overly worry about whether our variable is distributed normally. Since we are using the sample standard deviation for our estimate of the standard error, we will use a t-value for our critical value Test Statistic (z* or t*) t* = (69.22 - 65)/1.017 Rejection Region t.01/2, 62 d.f. = - 2.657 and 2.657 Calculation of Test Statistic t* = 4.151 Comparison of Test Statistics with Rejection Region t* > |t.01/2, 62 d.f.| 4.151 > 2.657. We can reject Ho: μ = 65; There is evidence to suggest people are speeding.

GE6

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