Question

(-76.25 Points] DETAILS BBBASICSTATSACC 8.1.018.MI. MY NOTES ASK YOUR TEACHER What price do farmers get for their watermelon crops? In the third week of July, a random sample of 42 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. Assume that o is known to be $1.92 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 (in dollars). What is the margin of error in dollars)? (For each answer, enter a number. Round your answers to two decimal places.) lower limit क upper limit $ margin of error (6) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.25 for the mean price per 100 pounds of watermelon. (Enter a number. Round up to the nearest whole number.) farming regions (c) A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop in dollars). What is the margin of error (in dollars)? Hint: 1 ton is 2000 pounds. (For each answer, enter a number. Round your answers to two decimal places.) Tower limit S 6 upper limit margin of error S

Answer 1

I'm not quite sure if you are joking or not, but here is your answer:
(a) The point estimate for the population mean price (per 100 pounds) is x = $6.88. The standard deviation is known to be σ = $1.92. The sample size is n = 42. For a 90% confidence interval, the critical value is 1.645 (obtained from a t-distribution table with 41 degrees of freedom). The margin of error is:

Margin of error = Critical value x Standard error

Standard error = Standard deviation / sqrt(sample size) = 1.92 / sqrt(42) = 0.2968

Margin of error = 1.645 x 0.2968 = 0.4882 ≈ 0.49

The lower limit of the confidence interval is:

Lower limit = Point estimate - Margin of error = 6.88 - 0.49 = $6.39

The upper limit of the confidence interval is:

Upper limit = Point estimate + Margin of error = 6.88 + 0.49 = $7.37

Therefore, the 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop is $6.39 to $7.37. The margin of error is $0.49.

(b) The maximal error of estimate is E = 0.25. The confidence level is 90%. The standard deviation is known to be σ = $1.92. We need to find the sample size (n) that satisfies the following formula:

Margin of error = Critical value x Standard error

Standard error = Standard deviation / sqrt(sample size)

For a 90% confidence interval, the critical value is 1.645. Substituting the known values and solving for n, we get:

0.25 = 1.645 x (1.92 / sqrt(n))

sqrt(n) = (1.645 x 1.92) / 0.25

n = [(1.645 x 1.92) / 0.25]^2

n ≈ 113

Therefore, a sample size of 113 farming regions is necessary for a 90% confidence level with maximal error of estimate E = 0.25 for the mean price per 100 pounds of watermelon.

(c) The farm brings 15 tons of watermelon to market, which is equivalent to 30,000 pounds (since 1 ton = 2,000 pounds). The point estimate for the population mean cash value of this crop is unknown. We can use the confidence interval obtained in part (a) to estimate this interval. Multiplying the lower and upper limits by 300 (since 300 pounds of watermelon correspond to $6.88), we get:

Lower limit = 6.39 x 300 = $1917

Upper limit = 7.37 x 300 = $2211

Therefore, the 90% confidence interval for the population mean cash value of this crop is $1917 to $2211. The margin of error is:

Margin of error = (Upper limit - Lower limit) / 2 = (2211 - 1917) / 2 = $147