Question

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 45 farming regions gave a sample mean of = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.94 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     $  
upper limit     $  
margin of error     $  

(b)

Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.41 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.)
lower limit     $  
upper limit     $  
margin of error     $  

Answer 1

Answer:

Given That,

$\rightarrow$ In the third week of July,

$\rightarrow$ A random sample of 45 farming regions gave a sample mean of = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.94 per 100 pounds.

i.e,

Sample mean =$\bar{x}$ = $6.88

Sample size = n = 45

Population standard deviation = $\sigma$ = $1.94

(a).

For 90% confidence interval, $\alpha$ = 1-0.90= 0.10

$\alpha$/2 = 0.10/2= 0.05

From normal distribution table, the value of z having an area of ($\alpha$/2 = 0.05) in its upper tail, is given by:

Z_0.05= 1.645

Margin of Error:

MoE = Z0.05 Χσινη

MoE= 1.645 x 1.94/145

MoE = 0.476

90% confidence interval for the population mean price (per 100 pounds):

Lower Limit of 90% confidence interval = 6.88 — 0.476 = $6.40

Upper Limit of 90% confidence interval = 6.88 + 0.476 = $7.36
Margin of error = $0.476

(b).

Given margin of error = E = 0.41

Margin of Error:

E = 20.05 Xo/vn

0.41 = 1.645 x 1.94/vn

n= 55.08 = 55.1 = 55

Therefore n=55 (Rounding up)

Hence, sample size necessary = 55 farming regions .

(c).

As calculated in (a),

90% confidence interval for the population mean price (per 100 pounds):

Lower limit of 9096 confidence interval = 6.88 — 0.476 = $6.40

Upper Limit of 00% confidence interval = 6.88 + 0.476 = $7.36

15 tons of watermelon =15 X 2000 = 30000 pounds

Lower Limit of 90% confidence interval for population mean price per 100 pounds = $6.40

Hence, lower limit of mean cash value of 15 tons of crop:

30000/100 X 6.40 = $1921.29

Upper Limit of 90% confidence interval for population mean price per 100 pounds =$7.36

Hence, Upper limit of mean cash value of 15 tom of crop:

= 30100000 /87.36 = $2206.71
Margin of error:

= $142.71

2206.71-1921.29