Question

  

Show, organize, and label (SOL) your work. Use correct notation as done in class or the book. If you use a calculator function to do a calculation, write down the keys used. You do not have to write down calculator steps if you’re just using a calculator to do arithmetic calculations.

1. At the Shenandoah Valley Produce Auction (SVPA) a box contains 5 ghost pumpkins with the following weights: {5 lbs, 7 lbs, 8 lbs, 11 lbs, 14 lbs}. Let X = weight of ghost pumpkins in the box.

a) Fill in the missing cells in the table below. Calculate the mean and standard deviation for all 3 columns. Note: You should use the population standard deviation formula.

	x	y=x-µ­X	z=(x-μ­X)/σX
	5 lbs	- 4 = 5 - 9
	7 lbs
	8 lbs
	11 lbs
	14 lbs
Mean μ	9 lbs
SD σ

b) How did transforming the original data (X) by subtracting the mean of the original data change the mean? In other words, what is the mean of Y? How did dividing by the standard deviation change the standard deviation of the original data? In other words, what is the standard deviation of Z?

c) Does the variable Z have a standard normal distribution? Why did you answer the way you did? Think about what properties a standard normal distribution has.

d) What is the probability that X is greater than 7 if a pumpkin is randomly chosen from the box (should be very simple to answer)? How does this compare to the probability that the weight of a randomly chosen pumpkin has a Z-score greater than the z-score of 7 (again easy to answer)?

2.. The SVPA sells a box of 6 Blue Hubbard pumpkins. The mean weight of all the pumpkins in the box is 14.5 lbs. The table below shows the distribution of the sample mean weight if 3 pumpkins are selected randomly from the box.

Sample Mean (lbs)	9	11	12	13	14	15	16	17	18	20
Probability	0.1	0.1	0.05	0.15	0.1	0.1	0.15	0.05	0.1	0.1

a. Suppose the three pumpkins selected are 3lbs, 9 lbs and 21 lbs. Construct a 60% confidence interval on the population mean using this sample. Explain how you figured out the margin of error.

b. Interpret the 60% confidence interval found in part a. Give the full interpretation and use the context of the problem

c. The SVPA claims that based on years of data the mean weight of the pumpkins in the box is 14.5 lbs. Does the 60% confidence interval from part a contradict this claim?

d. What is the probability the mean weight of the pumpkins in the box is in the 60% confidence interval found in part a. Explain.

e. For the hypotheses H₀: μ = 14.5 lbs and H₁: μ ≠ 14.5 lbs on the mean weight of the pumpkins in the box, what would the p-value be for the hypothesis test using the sample from part a. Use correct notation. Explain what the p-value represents within the context of the problem.

f. For the hypothesis test in part e, what would be the lowest significance level at which we would reject the null hypothesis? Explain.

3. Giant pumpkins frequently don’t contain seeds when cut open because of excessive in-breeding. This is unfortunate since seeds can sell for $30 each or more. In a sample of 50 giant pumpkins, 23 did not contain seeds.

a) What is the point estimate for the proportion of giant pumpkins that don’t have seeds?

b) Show that the necessary conditions are satisfied for the sample proportion to have an approximately normal distribution.

c) Using the sample above an obstreperous giant pumpkin researcher wants to use the unusual confidence level of 75% to create a confidence interval for the true proportion of giant pumpkins not containing seeds

                i) What would be the standard error for the sample proportion? Use p-hat as an estimate of p.

                ii) What is the critical value for this confidence interval?

                iii) What is the margin of error?

                iv) What is the75% confidence interval?

d) Interpret the 75% confidence interval within the context of the problem.

e) What sample size would be necessary to produce a margin of error equal to 5 percentage points assuming the true proportion is equal to the sample proportion from above?

Answer 1

Question 1 - solved

a.
Mean and standard deviation calculation

b) How did transforming the original data (X) by subtracting the mean of the original data change the mean? In other words, what is the mean of Y? How did dividing by the standard deviation change the standard deviation of the original data? In other words, what is the standard deviation of Z?

By subtracting the mean of the original data with mean we are only changing the scale of the data. The mean of the transformed data is 0.
By dividing the standard deviation we trying to find out how far the point is from the new mean.
The standard deviation of the Z is 1.

c) Does the variable Z have a standard normal distribution? Why did you answer the way you did? Think about what properties a standard normal distribution has.
Yes the variable Z has a standard normal distribution . A normal distribution has a mean of 0 and standard deviation of 1.

d) What is the probability that X is greater than 7 if a pumpkin is randomly chosen from the box (should be very simple to answer)? How does this compare to the probability that the weight of a randomly chosen pumpkin has a Z-score greater than the z-score of 7 (again easy to answer)?