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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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