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-bar = $6.88 per 100 pounds of watermelon. assume that s is known to be $1.92 per 100 pounds?

Answer 1

Given that a random sample of 40 farmers gave a sample mean of $\bar{x}=\ 6.88$ received by farmers per 100 pound of watermelon. Assume that σ is known to be $1.92 per 100 pounds.

Part A:

The 90% confidence interval for the population mean price (per 100lbs) that farmers get for their watermelon crop is given by:

$90\% \ C.I.=\bar{x}\pm1.645 \frac{\sigma}{\sqrt{n}} \\ \\ =6.88\pm1.645 \frac{1.92}{\sqrt{40}} =6.88\pm1.645(0.3036) \\ \\ =6.88\pm0.5=\bold{(\ 6.38,\ 7.38)}$

The margin of error is $0.50

The 99% confidence interval for the population mean price (per 100lbs) that farmers get for their watermelon crop is given by:

$99\% \ C.I.=\bar{x}\pm2.575 \frac{\sigma}{\sqrt{n}} \\ \\ =6.88\pm2.575 \frac{1.92}{\sqrt{40}} =6.88\pm2.575(0.3036) \\ \\ =6.88\pm0.78=\bold{(\ 6.10,\ 7.66)}$

A longer confidence interval should be expected for a higher confidence, because the longer the interval, the more likely it will contain the true population mean.


Part B

Recall that 300 times 100 lbs = 15 tons.

The 90% confidence interval for the population mean price (per 15 tons) that farmers get for their watermelon crop is given by:

$90\% \ C.I.=(6.38\times300,7.38\times300)=\bold{(\ 1,914,\ 2,214}$

The margin of error is $0.50 x 300 = $150