Given that a random sample of 40 farmers gave a sample mean of [tex]\bar{x}=\$6.88[/tex] 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:
[tex]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)}[/tex]
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:
[tex]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)}[/tex]
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:
[tex]90\%
\ C.I.=(6.38\times300,7.38\times300)=\bold{(\$1,914,\$2,214}[/tex]
The margin of error is $0.50 x 300 = $150
