Question

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 $1.98 per 100 pounds.
A farm brings 15 tons of watermelon to the 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

Answer:

a. [6.6350,7.3950]

b. ME=0.5150

Step-by-step explanation:

a. Given that n=40, $\bar x=6.88$ and that:$z_{\alpha/2}=z_{0.05}=1.645$

The required 90% confidence interval can be calculated as:

$\bar x\pm(margin \ of \ error)\\\\\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\\\6.88\pm(1.645\times \frac{1.98}{\sqrt{40}})\\\\=[6.3650,7.3950]$

Hence, the 90% confidence interval for the population mean cash value of this crop is [6.6350,7.3950]

b. The margin of error at 90% confidence interval is calculated as:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\\\=(1.645\times \frac{1.98}{\sqrt{40}})\\\\=0.5150$

Hence, the margin of error is 0.5150