Question

Question 1: On your shelf, you have four books that you are planning to read in the near future. Two are fictional works, containing 212 and 379 pages, respectively, and the other two are nonfiction, with 350 and 575 pages, respectively.

1. (a) Compute the population mean μ and standard deviation σof the (parent) population of number of pages of the N = 4 books.

2. (b) Suppose that you select a simple random sample of n = 2 books from the four (i.e., a random sample drawn without replacement) to take on a one-week ski trip (in case you injure yourself). Let X̄denote the sample mean number of pages for the two books selected. Obtain the sampling distribution of X̄.

3. (c) Determine the mean μX̄ and the standard deviation σX̄of the sampling distribution of X̄ and verify that μX̄ = μ and σ2̄ = (σ2/n)[(N − n)/(N − 1)].

Question 2: To avoid difficulties with the federal or provincial and local consumer protection agencies, a beverage bottler must make reasonably certain that 355 millilitre (mL) bottles actually contain 355 mL of beverage. To determine whether a bottling machine is working satisfac- torily, one bottler randomly samples n = 10 bottles per hour and measures the amount of beverage in each bottle. The sample mean X̄ of the 10 fill measurements is used to decide whether to readjust the amount of bever- age delivered per bottle by the filling machine. If records show that the amount of fill per bottle is normally distributed, with a standard deviation of σ = 5.91 mL, and if the bottling machine is set to deliver a mean fill per bottle of μ = 357.8 mL, what is the probability that the sample mean amount of beverage X̄ of the next 10 test bottles is less than 355 mL?

Question 3: The weight of a randomly selected bag of fertilizer has a mean of 50 lb and a standard deviation of 1 lb. What is the approximate probability that a random sample of 100 bags of fertilizer will weigh in total more than 5025 lb?

Question 4: Suppose that a particular candidate for public office is in fact favoured by 48% of all registered voters in the district. A polling organization will take a random sample of 500 voters and will usepˆ, the sample proportion of registered voters in favour of the candidate, to estimate p = .48. What is the approximate probability that pˆ will be greater than .5, causing the polling organization to incorrectly predict the result of the upcoming election?

Question 5:

1. (a) Why is an unbiased statistic generally preferred over a biased statistic for estimating a population parameter?

2. (b) Does unbiasedness alone guarantee that the estimate will be close to the true parameter value? Explain.

3. (c) Under what circumstances might you choose a biased statistic over an unbiased statistic if two such statistics are available for estimating a population parameter?

Question 6: Here are the weights in kilograms (kg) of a random sample of n = 24 male runners

67.8 61.9 63.0 53.1 62.3 59.7 55.4 58.9 60.9 69.2 63.7 68.3 64.7 65.6 56.0 57.8 66.0 62.9 53.6 65.0 55.8 60.4 69.3 61.7

Suppose that male runner weight is normally distributed, with a standard deviation of σ = 4.5 kg. Compute a 95% confidence interval for the popula- tion mean weight μ of a male runner. Are you quite sure that μ is less than 65 kg? Why?

Answer 1

1.

a) Population mean , = (212+379+350+575)/4 = 379

Population standard deviation , = {(212-379)^2 + (379-379)^2 + (350-379)^2 + (575-379)^2 )} / 4 = 129.5627

b) is the mean of the no. of pages of two books selected without replacement. The values of will be (212+379)/2 , (212+350)/2 , (212+575)/2 , (379+350/)/2 , (379+575)/2 , (350+575)/2 which is 295.5 , 281 , 393.5 , 364.5 , 477 , 462.5 respectively.

c) Mean of , = (295.5 + 281 + 393.5 + 364.5 + 477 + 462.5)/6 = 379

Therefore mean of , ​​​​​​​  is equal to the mean of the population .

Standard deviation of , = {(295.5-379)^2 + (281-379)^2 + (393.5-379)^2 + (364.5-379)^2 + (477-379)^2) + (462.5-379)^2)} / 6 = 74.8030

Variance of , ² = (74.8030)^2 = 5595.4977

Consider, (²/n)[(N-n)/(N-1)] = {(129.5627)^2/2} * { (4-2)/(4-1)} = 5595.49 = ²

Hence ² = (²/n)[(N-n)/(N-1)]

2)