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Self- assessment exam D

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Self- assessment exam D – p-value approach ..Name: _____________________________________________________Problem 1.When Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 428 green peas and 152 yellow peas. Find a 90% confidence interval estimate of the percentage yellow peas.a) Give the following information if they can be obtained from the information of the problem.______________,pˆ_______,p,peas________,peas______speasyellow_______x,peas________x,peas__________,peas_______nb) Write the formula appropriate for constructing the required confidence interval and substitute the values in the formula.answer: __________________________________________________c) We conclude with a 90% of confidence that the true percentage of yellow peas is between _______% and __________% . The error of estimation is ___________%Problem 2 – An economist wants to estimate the true mean income for the first year of work for college graduates who have had the profound wisdom to take a statistics course. A sample of 400such incomes shows a mean of $35000 with a standard deviation of $6000. Construct a 99% confidence interval for the true mean income for the first year of work of college graduates who have had the profound wisdom to take a statistics course.a) Give the following information if they can be obtained from the information of the problem.______________,______,s________,x_________________,n b) Write the formula appropriate for constructing the required confidence interval and substitute the values in the formula.answer: __________________________________________________c) We conclude with a 99% of confidence that the true mean income of college graduates is between $___________ and $______________ . The error of estimation is $____________-1-Problem 3 – An economist wants to estimate the true mean income for the first year of work for college graduates who have had the profound wisdom to take a statistic course. How many such incomes must be found if we want to be 99% confident that the sample mean is within $500 of the true population mean? Assume that a previous study has revealed that for such incomes.$6000a) Give the following information if they can be obtained from the information of the problem._______estimationofError______,________,_______,n b) Write the formula appropriate for determining how many incomes must be foun.Answer: ___________________answer: __________________________________________________c) __________incomes must be found if we want to be 99% confident that the sample mean is within $500 of the true population mean? -2-Problem 4: When Gregor Mendel conducted his famous hybridization experiments with peas, one such experiment resulted in offspring consisting of 428 peas with green pods and 152 with yellow pods. According to Mendel’s theory, ¼=0.25=25% of the offspring peas should have yellow pods. Use a 0.05 significance level with the p-value method to test Mendel’s claim that the proportion of peas with yellow pods is equal to ¼.Step 1. Is this a hypothesis about p or ? ____________________,pˆ_____,x___,___,s___,x______,n ______________:H__________:H10Step 2. Type of test – select: _____________ a) two-tailed test b) right-tailed test c) left –tailed test ____________Step 3. _________Step 4. Find the p-value – Show the work. p –value = __________Step 5. Compare the p-value with . Select: _________ a) p-value <  b) p-value > Step 6. Conclusion. Select: _______________________a) We reject the null hypothesis b) We fail to reject the null hypothesisSelect: ____________________a) There is not sufficient evidence to warrant rejection at a 0.05 level of the claim that ¼ of the offspring peas have yellow pods.b) We have statistical evidence to reject Mendel’s claim that ¼ of the offspring peas should have yellow pods.c) Mendel’s claim that ¼ of the offspring peas should have yellow pods is correct -3-Problem 5. A Tobacco Company advertised that its best selling non-filtered cigarettes contain at most 40 mg of nicotine, but Consumer Advocate magazine ran 10 randomly selected cigarettes and found the amounts (in mg) shown in the accompanying list. x 47.3 39.3 40.3 38.3 46.3 43.3 42.3 49.3 40.3 46.3It’s a serious matter to charge that the company advertising is wrong, so the magazine editor chooses a significance level of 010. in testing her believe that the mean nicotine content is greater than 40 mg. Using a 0.01 significance level, test the editor’s believe that the mean is greater than 40mg.a) Find the mean and standard deviation of the given sample (you can use a calculator) mean = __________ mg, standard deviation = __________b) Complete if possible ___.________,____,.s__,.x___________,n 01080334310 c) Test the hypothesis.Step 1. Is this a hypothesis about p or ? __about____________________,______,s________,x_________________,n Step 1. Is this a hypothesis about p or ? ______________________:H__________:H10Step 2. Type of test – select: _____________ a) two-tailed test b) right-tailed test c) left –tailed test ____________Step 3. _________Step 4. Find the p-value – Show the work. p –value = __________Step 5. Compare the p-value with . Select: _________ a) p-value <  b) p-value > Step 6. Conclusion. Select: _______________________a) We reject the null hypothesis b) We fail to reject the null hypothesisSelect: ____________________a) Statistically at a 0.01 level of significance we have evidence to agree with the editor’s believe that the mean is greater than 40mg.b) ) Statistically at a 0.01 level of significance we have evidence to agree with the Tobacco Company advertised that its best selling non-filtered cigarettes contain at most 40 mg of


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