Statistics and Probability

Question

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. Assume that 𝜎 is known to be $1.94 per 100 pounds.

(a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop. What is the margin of error? (Round your answers to two decimal places.)

(b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.27 for the mean price per 100 pounds of watermelon. (Round up to the nearest whole number.)

(c) A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop. What is the margin of error? Hint: 1 ton is 2000 pounds. (Round your answers to two decimal places.)

Answer 1

Part a:

The general form of a two-sided confidence interval is the following:

(best estimate or sample mean) +/- (z_crit) * (sample mean standard deviation)

where z_crit is a value in the z-table that corresponds to the (1-alpha) percentile. For a 90% confidence interval (alpha = 0.1), z_crit is approximately 1.28. Try to figure out how I got this value.

Next, the sample mean standard deviation is equal to the standard deviation of one sample divided by the square root of the number of samples you have. From the problem statement, the standard deviation of one sample is given as 1.94 and the number of samples is equal to 40. So, the sample mean standard deviation is 1.94/sqrt(40), or approximately 0.307.

With all of these pieces, you can calculate the margin of error = z_crit * (sample mean standard deviation).

Part b:

For this question, you have the margin of error, z_crit, and the standard deviation of one sample. The only unknown is n. You can follow a similar procedure in part a to determine n.

It may also be worthwhile to understand why it is important to determine sample size from a practical standpoint.

Part c:

You will need to determine n via unit conversions (i.e. what must n be for 15 tons of watermelon is n = 1 corresponds to 100 pounds of watermelon). Then, follow the same procedures as parts a and b to determine the margin of error.