(e) Enough proof at the 1% level to support the claim that the mean repair expenditure for damaged Microwave is below $100

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1 Individual Assignment #3 Section 7.3 #23 Testing Claims (a) write the claim mathematically and identify Ho and Ha. (b) find the critical value(s) and identify the rejection region(s). (c) find the standardized test statistic. (d) decide wether to reject or fail to reject the null hypothesis. (e) interpret the decision in the context of the original claim. For each claim, assume the population is normally distributed. Microwave Repair Costs A microwave oven repairer says that the mean repair cost for damaged microwave ovens is less than $100. You work for the repairer and want to test this claim. You find that a random sample of five microwave ovens has a mean repair cost of $75 and a standard deviation of $ At α = 0.01, do you have enough evidence to support the repairer s claim? (a) H 0 : 100 H a : < 100 (claim) (b) Critical Value: t 0 = Rejection Zone: t < (c) Standardized Test Statistic: (d) Reject H 0 (e) Enough proof at the 1% level to support the claim that the mean repair expenditure for damaged Microwave is below $100 #37 Deciding on a Distribution In Exercises 37 and 38, decide whether you should use a normal sampling distribution or a t-sampling distribution to perform the hypothesis test. Justify your decision then use the distribution to test the claim. Write a short paragraph about the results of the test and what you can conclude about the claim. Gas Mileage A car company says that the mean gas mileage for its luxury sedan is at least 21 miles per gallon (mpg). You believe the claim is incorrect and find that a random sample of five cars has a mean gas mileage of 19 mpg and a standard deviation of 4 mpg. Assume the gas Individual Assignment 3 Page 1

2 mileage of all the company s luxury sedans is normally distributed. At α = 0.05, test the company s claim. a. Use the t distribution because population is normal b. n < 30, and o is unknown c. Fail to Reject H 0. d. There is enough proof at the 5% level to refuse the claim that the mean gas mileage for luxury sedan is minimum 21 gal for each gallon Section 7.4 #10 Testing Claims In Exercises 9-14, (a) write the claim mathematically and identify Ho and Ha, (b) find the critical value(s) and identify the rejection region(s), (c) find the standardized test statistic, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Do You Eat Breakfast? A medical research estimates that no more than 55% of U.S. adults eat breakfast every day. In a random sample of 250 U.S. adults, 56.4 say they eat breakfast every day. At α = 0.01, is there enough evidence to reject the researcher s claim? (a) H 0 : 56.4 H a : < 56.4 (Claim) (b) Critical Value: t0 = t < (c) Standardized Test Statistic: 1 z = 1 (d) Agree to H0 (e) Enough proof To Support 55% (or.55) of U.S. adults eat breakfast. #15 Free Samples In Exercises 15 and 16, use the graph, which shows what adults think about the effectiveness of free samples. Individual Assignment 3 Page 2

3 Do Free Samples Work? You interview a random sample of 50 adults. The results of the survey show that 48% of the adults said they were more likely to buy a product when there are free samples. At α = 0.05, can you reject the claim that at least 52% of the adults are more likely to buy a product when there are free samples? Not succeed to Reject H 0. There isn t sufficient proof at the 5% level to refuse the claim that minimum 52% of the adults more prone to purchase an item when there are free samples. Section 7.5 #17 Testing Claims In Exercises 15-24, (a) write the claim mathematically and identify Ho and Ha, (b) find the critical value(s) and identify the rejection region(s), (c) use the X 2 -test to find the standardized test statistic, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. For each claim, assume the population is normally distributed. Physical Science Assessment Tests On a physical science assessment test, the scores of a random sample of 22 eighth grade students have a standard deviation of This result prompts a test administrator to claim that the standard deviation for eighth graders on the examination is less than 29. At α = 0.10, is there enough evidence to support the administrator s claim? (a) Ho: o 29; Ha: o < 29 (Claim) Individual Assignment 3 Page 3

4 Critical Value: X 2/0 = ; (b) Rejection Zone: : X 2 < (c) Standardized Test Statistic: X 2 = (d) Fail to reject Ho (e) Inadequate proof at the 10% level of significance to support the claim that the standard deviation for the eighth graders on the exam is below 29. #24 Salaries An employment information service says that the standard deviation of the annual salaries for public relations managers is at least $14,500. The annual salaries for 18 randomly chosen public relations managers are listed. At α = 0.10, can you reject the claim? 47,517 60,217 39,177 61,744 79,422 70,770 60,549 60,263 72,939 72,372 75,014 59,164 44,811 65,419 61,310 90,433 44,185 41,805 (a) H 0 : $14,500 (claim) H a : $14,500 (b) Critical Value: t0 = Rejection Zone: t < (c) Standardized Test Statistic: X 2 = -4 (d) Reject the statement because the actual Standard Deviation is $14, Individual Assignment 3 Page 4

5 Chapter 6 Case Study #1 Use the sample to find a point estimate for the mean shoulder height of (a) male bears. (b) female bears. (a) The point estimate for the mean shoulder height of male bears is (b) The point estimate for the mean shoulder height of female bears is #2 Find the standard deviation of the sample of shoulder heights for the (a) male bears. (b) female bears. (a) The standard deviation of the sample of shoulder heights for the male bears is (b) The standard deviation of the sample of shoulder heights for the female bears is 7.59 #3 Use the sample to construct a 95% confidence interval for the mean shoulder height of (a) male bears. (b) female bears. (a) The 95% confidence interval for the mean shoulder height of male Individual Assignment 3 Page 5

6 bears ,83.48 (b) The 95% confidence interval for the mean shoulder height of female bears , #4 Use the sample to construct a 95% confidence interval for the mean shoulder height of all bears in the study. How do your results differ from those in Exercise 3? Explain. The 95% confidence interval for the mean shoulder height of all bears ,80.05 The major difference is that the all the measurements are used, both male and female to come up with the standard deviation of all the samples. Having a bigger sample appears to decrease the margin of error as is shown in the reading. Chapter 7 Case Study pg 363 #1 Complete the hypothesis test for all adults (men and women by performing the following steps. Use a level of significance of α = (a) Sketch the sampling distribution. (b) Determine the critical values and add them to your sketch. (c) Determine the regions and shade them in your sketch. (d) Find the standardized test statistic. Add it to your sketch. Individual Assignment 3 Page 6

7 (e) Make a decision to reject or fail to reject the null hypothesis. (f) Interpret the decision in the context of the original claim. (a) (b) Significance level = 0.05 (sample mean should have under 5% chance of taking place at 0.05 Zc = ±1.96 (c) Individual Assignment 3 Page 7

8 Rejection area is displayed in black. The two temperatures corresponding to Zc are displayed as and (d) Z value for sample mean is Z = Xbar μ / (s/sqrt(n)) Z = (e) As Z < Zc we can refuse the claim. (f) The original claim was that the mean value of body temperature is Depending on the study sample we can deduce that the usual body temperature is under #2 If you lower the level of significance to α = 0.01, does your decision change? Explain your reasoning. Individual Assignment 3 Page 8

9 At significance level of 0.01 we have Zc = ± Again as Z< Zc we can still refuse the claim. #3 Test the hypothesis that the mean temperature of men is 98.6 F. What can you conclude at a level of significance of α = 0.01? mean= std dev= standardized test statistic= critical values= ±2.576 As Z< Zc we can refuse the claim that the mean temperature of mean is 98.6 #4 Test the hypothesis that the mean temperature of women is 98.6 F. What can you conclude at a level of significance of α = 0.01? mean= std dev= standardized test statistic= critical values= ±2.576 As Z > Zc and is not in the refusal area we must accept the null hypothesis and claim that the mean temperature of ladies is 98.6 #5 Use the sample of 130 temperatures to form a 99% confidence interval for the mean body temperature of adult humans. Individual Assignment 3 Page 9

10 E = Zc * s/(sqrt(n)) E = ±2.574 * 0.73/SQRT(130) E = ± With 99% confidence we might say that the mean value of human body temperature is between and #6 The conventional normal body temperature was established by Carl Wunderlich over 100 years ago. What, in Wunderlich s sampling procedure, do you think might have led him to an incorrect conclusion? Strictly from a sampling method point of view, from the data analysis in parts 3 and 4 we can see that in case the sample contained considerably more ladies compared to guys, it would be more likely to create a somewhat higher mean value of body temperature. Individual Assignment 3 Page 10

Name: Date: Use the following to answer questions 3-4:

Name: Date: Use the following to answer questions 3-4: Name: Date: 1. Determine whether each of the following statements is true or false. A) The margin of error for a 95% confidence interval for the mean increases as the sample size increases. B) The margin