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STAT 200 Guided Exercise 6 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 the following. Please note that it is an algebraic manipulation of the confidence interval formula to solve for n based on a desired plus or minus bound of error of .8 mg/l. a. What is the minimum sample size you would need? Assume the water readings are approximately normal with σ = 5 mg/L α/2 = .025 since (1-.95) =.05; z= 1.96; B = .8 mg/l N = 151 as a minimum sample size b. Check you answer by calculating the confidence interval for this problem (the Bound of Error) using the sample size you calculated. 2. A larger supermarket began to get complaints from customers about the quantity of chips in a 16ounce bag of a particular brand of potato chips. The chain decided to test the following hypothesis concerning µ, the mean weight (in ounces) of a bag of potato chips in the next shipment of chips received from their supplier. Ho: µ = 16 Ha: µ < 16 If there is evidence that µ < 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. 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. b. 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 shipmentsThey would be concerned with Type II Error. d. Which type of error would the supplier view as more serious? 1 The supplier would rather have their product accepted – They would be concerned with Type I Error, rejecting shipments when the weight was ok. 3. Current technology uses X-rays and lasers for inspection of solder-joints on printed circuit boards. A current manufacturer claims its product can inspect on average at least 10 solder joints per second. A potential buyer tested the equipment on 27 different printed circuit boards (PCB). The sample mean and standard deviation are given below. We will assume that this distribution is approximately normal. This is a small sample hypothesis test of a mean. Mean = 10.54 s = 1.61 n = 27 a. Calculate a 95% confidence interval for this sample estimate. Standard Error = 1.61/ (27).5 = .3098; t.05/2, 26 d.f. = 2.056 10.54 ± 2.056(.3098) 10.54 ± .637 9.903 to 11.177 b. Test to see if the sample data is different from the manufacturer’s claim. 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=10 Alternative Hypothesis Ha : µ≠10 Two-tailed Test Assumptions of Test n=27 is a small sample. So I use the t-distribution, with α=0.05/2 and 27-1=26 d.f. We assume the distribution in the population is approximately normal. Test Statistic (z* or t*) t*=(10.54-10.0)/.3098 Rejection Region t.05/2,26 d.f.=2.056 Calculation of Test Statistic t*=1.743 Comparison of Test Statistics with Rejection Region -2.056<t*<2.056 Cannot reject H0 2 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. a. What is the proportion for our sample? Est. 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 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 p = .80 Alternative Hypothesis p < .80 Assumptions of Test use normal approximation to binomial Test Statistic (z* or t*) z*= (.6778 - .80)/.0422 Rejection Region z.05 = -1.645 Calculation of Test Statistic z* = -2.90 Comparison of Test Statistics with Rejection Region z* < z.05 ; 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? 3 With a z* = -2.90 the p-value is: .5 - .4981 = .0019 This is much lower than α = .05, so we can reject the Null Hypothesis What if our alternative hypothesis was a two-tailed test, what would the p-value be? If this is a two-tailed test, p= .0019*2 = .0038 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. 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. HYPOTHESIS TEST Null Hypothesis p = .50 Alternative Hypothesis p >.50 Assumptions of Test use normal approximation to binomial Test Statistic (z* or t*) z*= (.467 - .50)/SQRT[.5*.5/150] Rejection Region z.01 = 2.33 Calculation of Test Statistic z* = -.8083 Comparison of Test Statistics with Rejection Region z* < z.01 Cannot Reject Ho: p = .50 We do not have evidence that the sample proportion is any different from .50. b. Calculate the p-value for this test. This one is a little tricky, since we expected the est. p to be above .5 when in fact it was below .5 different from .50. (The TT practitioners did worse than chance). So the p-value is bases on z* = -.81 Lookup in table .2910 Plus .5 = .7910 4 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 Excel output from the descriptive statistics of the speed of drivers and the Stem and Leaf Plot from JMP. Briefly (one paragraph) describe the data using the summary statistics. The mean speed for the 63 motorists was 69.2 mph with a range of 39. The highest speed was 95 and the lowest was 56. The distribution is approximately mound shaped and symmetrical and the mean, median, and mode are all very similar, however the mean is slightly higher than the median and is being pulled by a few extreme high speeds. The Stem and Leaf plot shows the distribution is skewed right toward some high values. There is a fair amount of variability in the data with a standard deviation of 8. b. Conduct a Hypothesis Test where the null hypothesis is that the average speed is 65 mph. You are free to decide if it should be a one or two tailed test. Use an α level of .01. HYPOTHESIS TEST Null Hypothesis µ = 65 Alternative Hypothesis µ > 65 Assumptions of Test N is large (63) so we can use a z-value Test Statistic (z* or t*) (69.22-65)/1.016 Rejection Region Z= 2.33 Calculation of Test Statistic Z* = 4.1535 Comparison of Test Statistics with Rejection Region Z* > z We can reject Ho: µ = 65; There is evidence to suggest people are speeding 5 7. Golf course designers are worried that the new equipment is making old courses obsolete. One designer says that courses need to be built with the expectation that players will be able to drive the ball an average of 250 yards or more. A sample of 135 golfers is taken and they measured their driving distance: n=135 mean = 256.3 yards s = 43.4 yards a. Does the sample provide enough evidence to suggest that golfers are already hitting it farther than the 250 mark? Use α=.05 HYPOTHESIS TEST Null Hypothesis H0:µ= 250 Alternative Hypothesis Ha: µ > 250 one-tailed test, upper Assumptions of Test Large sample, normal Test Statistic (z* or t*) z* = (256.3-250)/(43.4/135.5) Rejection Region z =.05 = 1.645 Calculation of Test Statistic z* = 1.69 Comparison of Test Statistics with Rejection Region z* > za=.05 1.69 > 1.645 Reject H0: µ= 250 α b. Take the test statistic that you calculated (i.e., t* or z*) and find the probability associated with this test statistic value. If it is a z-value, the probability will be: .5 - p(z*) this is known as a p-value for this problem z* = 1.69 P(0<x<z*) = .4545 p= .5 - .4545 = .0455 Compare this to the alpha level set for the problem. α =.05 The p-value is less than .05, therefore I have a z* that is out in the rejection region When the p-value is less than α, I can reject the Null Hypothesis 6