Assignment-3

2B03 Assignment 3 Sampling Distributions and Statistical Inference (Chapters 5, 6 & 7) Matthew Musulin 400329990 Due Thursday November 4 2021 Instructions: You are to use R Markdown for generating your assignment output file. You begin with the R Markdown script downloaded from A2L, and need to pay attention to information provided via introductory material posted to A2L on working with R, R Markdown. Having added your answers to the R Markdown script, you then are to generate your output file using “Knit to PDF” and, when complete, upload both your R Markdown file and your PDF file to the appropriate folder on A2L. 1. Define the following terms in a sentence (or short paragraph) and state a formula if appropriate (this question is worth 5 marks). i. Hypothesis Test A test to choose between two conflicting hypotheses about the value of a population parameter ii. Estimator A function which tells us how to use a sample to estimate a population parameter iii. Interval Estimate Interval Estimation is the use of sample data to calculate an interval of possible values of an unknown population parameter iv. Efficiency When comparing two unbiased estimators, the estimator with the smallest variance is said to be efficient v. Consistency An estimator is consistent if its value gets ever closer to the parameter being estimated as the sample size increases 2. If the income in a community is normally distributed, with a mean of $38,000 and a standard deviation of $6,000, what maximum income does a member of the community have to earn in order to be in the bottom 5%? What is the maximum income one can have and still be in the middle 50% (this question is worth 4 marks)? mu = 38000 sd = 6000 z <- qnorm(. 05 ) bottom = z*sd + mu bottom ## [1] 28130.88 mu = 38000 sd = 6000 z <- qnorm(. 75 ) middle = z*sd + mu middle ## [1] 42046.94 3. Suppose that the number of hours per week of lost work due to illness in a certain automobile assembly plant is approximately normally distributed, with a mean of 40 hours and a standard deviation of 15 1

hours. For a given week, selected at random, what is the probability that (this question is worth 3 marks): i. The number of lost work hours will exceed 70 hours? P ( X > 70) = P ( Z > (70 − 40) / 15)) P ( Z > 2) = 0 . 0227501 ii. The number of lost work hours will be between 30 and 45 hours? P ( X > 30) − P ( X < 45) P ( Z > (30 − 40) / 15) − P ( Z < (45 − 40) / 15) P ( Z > − 0 . 667) − ( Z < 0 . 333) (1 − 0 . 25463) − 0 . 62930 = 0 . 11607 iii. The number of lost work hours will be exactly 50 hours? P ( X = 50) = 0 4. A senator claims that 58% of her constituents favour her voting policies over the past year. In a random sample of 50 of these people, the sample proportion of those favoured her voting policies was only 0.4. Is this enough evidence to make the senator’s claims strongly suspect? (Hint: Use a normal approximation to the binomial distribution then construct a confidence interval - this question is worth 2 marks). SE = sssssss pq n SE = qqqqqqq 0 . 4 ∗ 0 . 6 / 50 SE = 0 . 06928 CI = 0 . 4 + − 1 . 96 ∗ 0 . 06928 CI = [0 . 2642 , 0 . 5358] Because 0.58 doent lie within the interval, the senators claim is rejected. 5. I wish to estimate the proportion of defectives in a large production lot with plus or minus D = 0 . 02 of the true proportion, with a 99% level of confidence. From past experience it is believed that the true proportion of defectives is π = 0 . 02 . How large a sample must be used? (Hint: Use a normal approximation for the sample proportion ˆ P - this question is worth 2 marks). Z value for 99% level of confidence is 2.576 Margin of error, D = 0.02 n = ( Z/D ) 2 ∗ π ∗ (1 − π ) n = (2 . 576 / 0 . 02) 2 ∗ 0 . 02 ∗ (1 − 0 . 02) n = 325 . 153024 Therefore the sample size must be approximately 325 2