BCTC STATS EXAM 3 QUESTIONS WITH CORRECT DETAILED ANSWERS GRADED A+

BCTC STATS EXAM 3 QUESTIONS WITH CORRECT DETAILED ANSWERS GRADED A+ conf. = 90% --- 1.645 (1.645*67/3)^2 = 1349.705136 ~~~ 1350 - Answer-How many adults must be randomly selected to estimate the mean FICO (credit rating) score of working adults in a country? We want 90% confidence that the sample mean is within 3 points of the population mean, and the population standard deviation is 67. conf.= 99% --- 2.58 a) best point estimate = 153.64 lb b) 99% confidence interval estimate: 2.58*30.67 / sqrt(50) = 11. 153.64 +/- 11.= [142.45,164.83] - Answer-Using the simple random sample of weights of women from a data set, we obtain these sample statistics: n = 50 and x over-bar = 153.64lb. Research from other sources suggests that the population of weights of women has a standard deviation given by sigma = 30.67lb. conf. 90% -- 1.645 a) 1.645*8.3 / sqrt(40) = 2. 58.4 +/- 2. = [56.2,60.6] b) Based on the result, is it likely that the students' estimates have a mean that is reasonably close to sixty seconds? = Yes, because the confidence interval includes sixty seconds. - Answer-Randomly selected students participated in an experiment to test their ability to determine when one minute (or sixty seconds) has passed. Forty students yielded a sample mean of 58.4 seconds. Assuming that sigma = 8.3 seconds, construct and interpret a 90% confidence interval estimate of the population mean of all students. a) what is the 99% confidence interval for the mean? 2.58*6 / sqrt(14) = 4. 137 +/- 4. = [132.9,141.1] b) Ideally, what should the confidence interval in this situation be? Ideally, all the measurements would be the same, so there would not be an interval estimate. - Answer-When fourteen different second-year medical students measured the blood pressure of the same person, they obtained the results listed below. Assuming that the population standard deviation is known to be 6mmHg, construct and interpret a 99% confidence interval estimate of the population mean. [144, 121, 127, 145, 138, 148, 138, 138, 148, 124, 147, 145, 144, 137, 129, 131] - Answer-In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol (in mg/dL) have a mean of 4.9 and a standard deviation of 16.8. Complete parts (a) and (b) below. z= -2.45 - Answer-The claim is that the proportion of peas with yellow pods is equal to 0.25 (or 25%). The sample statistics from one experiment include 450 peas with 90 of them having yellow pods. z= -0.73 --- (0.24-0.25)/sqrt(0.25(1-0.25))/1005 p=0.2327 conclusion? There is not sufficient evidence to support the claim that less than 25% of adults have smoked within the past week. - Answer-In a poll of 1005 adults, it was found that 24% smoked cigarettes in the past week. Use a 0.05 significance level to test the claim that less than 25% of adults have smoked within the past week. Use this information to answer the following questions. red candy: 5/32 = 0.15625 z= 0.44 p= 0.6599 bc it is a two tailed p-value There is not sufficient evidence to warrant rejection of the claim that 13% of the candy maker's candies are red. - Answer-Test the claim that 13% of a candy maker's candies are red. Use a 0.05 significance level. Use the data in the table to the right to answer the following questions. test statistic: (0.8580-0.8565)/(0.0565/sqrt(27)) = 0.14 critical values = -1.96,1.96 final conclusion: Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the mean weight of jellybeans is equal to 0.8565g. - Answer-A sample of 27 blue jellybeans with a mean weight of 0.8580g was taken. Assume that sigma is known to be 0.0565g. Consider a hypothesis test that uses a 0.05 significance level to test the claim that the mean weight of all jellybeans is equal to 0.8550g (the weight necessary so that bags of jellybeans have the weight printed on the package). Assume the weight of jellybeans is normally distributed. z= -2.40 p= 0.0164 (it is two tailed so use the right tailed value and multiply by 2) final conclusion? Reject H0. There is sufficient evidence to warrant rejection of the claim that the mean is equal to 60sec. Does there appear to be an overall perception of 1 min that is relatively close to an actual minute? =it appears that the mean overall perception is not relatively close because there is sufficient evidence to warrant rejection of the claim. - Answer-Randomly selected statistics students participated in an experiment to test their ability to determine when 1 min (or 60 sec) has passed. Forty students yielded a sample mean of 56.4sec. Assuming that sigma= 9.5sec, use a 0.05 significance level to test the claim that the population mean is equal to 60sec. Based on the result, does there appear to be an overall perception of 1 min that is relatively close to an actual minute? z= -1.19 critical value= -1.645 (no idea how) final conclusion: fail to reject H0. There is not sufficient evidence... - Answer-A simple random sample of 45 salaries of professional football coaches has a mean of $417,022The standard deviation of all salaries of professional football coaches is $467 comma 618. Use a 0.05 significance level to test the claim that the mean salary of a professional football coach is less than $500,000 t = 2.71 (calculate like z value) p = 0.005 conclusion: Reject H0, there is sufficient evidence... What do the results suggest about the advice given in the manual? The results suggest that the advice of writing a song that must be no longer than 210 seconds is not sound advice. - Answer-In a manual on how to have a number one song, it is stated that a song must be no longer than 210 seconds. A simple random sample of 40 current hit songs results in a mean length of 232.9sec and a standard deviation of 53.36sec. Use a 0.05 significance level and the accompanying Minitab display to test the claim that the sample is from a population of songs with a mean greater than 210sec. calculator says -1.88, but it is 1.88 - Answer-Find the critical value z Subscript alpha divided by 2 that corresponds to the given confidence level. 94% (1.96)^2*(0.5(1-0.5)) / (0.04)^2 = ~601 - Answer-Use the given data to find the minimum sample size required to estimate a population proportion or percentage. margin of error:0.04, confid.:95% p^ and q^ unknown sqrt{(0.55(1-0.55))/480} = 0. 0.55 +/- (2.58*0.) = [0.491p0.609] np, it is not significantly different than 0.5 - Answer-A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 480 babies were born, and 264 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born. Based on the result, does the method appear to be effective? 4935 / 10000 = 0.4935 conf. = 95% --- 1.96 a) 1.96* sqrt{(0.4935(1-0.4935))/10000} = 0. 0.4935 +/- 0. = [0.484, 0.503] b) no, because the proportion could easily equal 0.5. The interval is not less than 0.5 the week before the holiday. - Answer-In the week before and the week after a holiday, there were 10,000 total deaths, and 4935 of them occurred in the week before the holiday. a)Construct a 95 %confidence interval estimate of the proportion of deaths in the week before the holiday to the total deaths in the week before and the week after the holiday. b) Based on the result, does there appear to be any indication that people can temporarily postpone their death to survive the holiday? p= 172/(172+445)=0.28 conf.= 90% ---1.645 a) 1.645* sqrt{(0.28(1-0.28))/617} = 0. 0.28 +/- 0. = [0.250,0.310] b) No, the confidence interval includes 0.25, so the true percentage could easily equal 25% - Answer-A genetic experiment with peas resulted in one sample of offspring that consisted of 445 green peas and 172 yellow peas. a. Construct a 90% confidence interval to estimate of the percentage of yellow peas. b. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations? E) neither normal nor t distribution applies - Answer-Do one of the following, as appropriate. (a) Find the critical value z Subscript alpha divided by 2, (b) find the critical value t Subscript alpha divided by 2, (c) state that neither the normal nor the t distribution applies. Confidence level 99%; n equals 17;sigma is known; population appears to be very skewed B) There is sufficient evidence to support the claim that the proportion of male golfers is less than 0.6. - Answer-State the final conclusion in simple nontechnical terms. Original claim: The proportion of male golfers is less than 0.6. Initial conclusion: Reject the null hypothesis. c) reject H0 since the test statistic 25.500 is greater than the critical value 2.896 - Answer-The heights were measured for nine supermodels. They have a mean of 68.7in. and a standard deviation of 0.6in. Use the traditional method and a 0.01 significance level to test the claim that supermodels have heights with a mean that is greater than the mean of 63.6 in. for women from the general population.