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slamgirl [31]
3 years ago
10

What price do farmers get for their watermelon crops? in the third week of july, a random sample of 40 farming regions gave a sa

mple mean of x-bar = $6.88 per 100 pounds of watermelon. assume that s is known to be $1.92 per 100 pounds?
Mathematics
1 answer:
Kobotan [32]3 years ago
6 0
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
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