STAT 308 Homework #8 Due Tuesday 14 November Fall 2017 URI NOTE: All questions should be handed in. Selected questions would be graded. 1. Do Exercises 9.15, 9.17 & 9.20 on page 374 in the textbook. 2. Do Exercise 9.85 on page 397 in the textbook. 3. Beer Drinking. The mean annual consumption of beer per person in the US is 22.0 gallons . A random sample of 300 W ashington D.C. residents yielded a mean annual beer consumption of 27.8 gallons. At the 10% significance level, do the data provide sufficient evidence to conclude that the mean annual consumption of beer per person for the nation's capital differs from the national mean? Assume that the standard deviation of annual beer consumption for W ashington D.C. residents is 55 gallons. 4. Do Exercise 9.63, 9.64 & 9.66 on page 387 in the textbook. In each one of the problems answer part (a) only. Also, give the excel command needed to calculate each of the p-values for each of the questions. No need to hand in any excel output, just write by hand the command you would use to find the p-values for each of the questions. No excel commands, no credit. 5. Do Exercise 9.83 on page 397 in the textbook. Use the p-value approach to hypothesis testing. Assume normality for the cadmium level in mushrooms. 6. Do Exercise 3 above but use the p-value approach to hypothesis testing. 7. Brewery Effluent and Crops. Because many industrial wastes contain nutrients that enhance crop growth, efforts are being made for environmental purposes to use such wastes on agricultural soils. Researchers studied the physico-chemical properties of effluent from Mohan Meakin Breweries Ltd. (MMBL), Ghazibad, UP, India, and \"... its effects on the physico-chemical characteristics of agricultural soil, seed germination pattern and the growth of two common crop plants.\" They assessed the impact of using different concentrations of the effluent: 25%, 50%, 75% and 100%. The following data, based on the results of the study, provide the percentages of limestone in the soil obtained by using 100% effluent: 2.41 2.60 2.31 2.51 2.54 2.51 2.28 2.42 2.72 2.70 Do the data provide sufficient evidence to conclude, at the 15% level of significance, that the mean available limestone in soil treated with 100% MMBL effluent exceeds 2.3%? That is, use "=0.15. Answer the question by writing the null and alternative hypothesis, test statistics value, critical value, draw acceptance and rejection regions, make a decision and write a conclusion. Also, calculate the exact p-value associated with the test. You need to use excel to get the critical value as well as the exact p-value (values not in t-table). Do not hand in any excel output but make sure that you write/give the excel commands you used to find the critical value as well as the p-value. No excel commands (by hand), no credit for the question. Exercises for recitations (11/07 - 11/13) ___________________________________________________________ Excel Instructions: The following functions applied to both, PCs and MACs. Working with the Standard Normal Distribution - (From homework 5) NORMSDIST(z) returns the probability that the observed value of a standard normal random variable will be less than or equal to z. That is, returns the area to the left of the given z-cores Syntax =NORMSDIST(z) where z is a z-score. NORMSINV(probability) returns the inverse of the standard normal cumulative distribution. That is, returns the z-score with an area to the left equal to probability Syntax =NORMSINV(probability) where probability is a probability corresponding to the normal distribution. Working with the t- distribution: TDIST returns the probability for the Student t-distribution where a numeric value x ($0) is a value of t for which the probability is to be computed. Syntax x df =TDIST(x,df,tails) where is the numeric value (non-negative) at which to evaluate the distribution is an integer indicating the number of degrees of freedom. tails If tails = 1, TDIST returns the area to the right of the non-negative x value. If tails = 2, TDIST returns the area to the right of non-negative x multiplied by two -two-tailed distribution-. TINV returns the t-value of the Student's t-distribution as a function of the probability and the degrees of freedom. Syntax =TINV(prob,df) where prob probability associated with the two-tailed Student's t-distribution (total area in both tails). If one-tailed then multiply " by 2 to get the correct tvalue. df integer indicating the number of degrees of freedom. _____________________________________________________________________ 1. (The wording in this paragraph was also given in the recitation exercises for assignment 7) As reported by the Department of Agriculture in Crop Production, the mean yield of oats for U.S. farms is 58.4 bushels per acre. A farmer wants to estimate his mean yield using an organic method. He uses the method on a random sample of 25 1-acre plots and obtained a mean of 61.49 and a standard deviation of 3.754 bushels. Assume yield is normally distributed. Assume now that the standard deviation is a population standard deviation. Does it appear that the farmer will get a mean yield different from the national average by using the organic method? Answer this question by carrying out a test of hypothesis. Use "=0.01. Give excel commands to find the critical value as well as the exact p-value associated with this test. 2. Biological Oxygen Demand (BOD) is an index of pollution that is monitored in the treated effluent of paper mills on a regular basis. From 43 determinations of BOD (in pounds per day) at a particular paper mill during the Spring of 1992, the mean and SD were found to be 3,246 and 757, respectively. The company has set the target that the mean BOD should be 3,000 pounds per day. Do the sample data indicate that the actual amount of BOD is significantly off the target? Use "=0.05. Use excel to find the exact critical value as well as the exact pvalue associated with this test. 3. Refer to problem above. Along with the determinations of BOD, the discharge of suspended solids (SS) was also monitored at the same site. The mean of the 43 determinations of SS was found to be 5,710 and the population standard deviation is assumed to be 1,720 pounds per day. Do these results strongly support the company's claim that the true mean SS is lower than 6,000 pounds per day? Use "=0.05. Use excel to find the critical value as well as the exact p-value associated with this test