What price do farmers get for their watermelon crops? In the third week of July, a random sample of 37 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. Assume that is known to be $1.86 per 100 pounds.
(a)
Find the margin of error for a 90% confidence interval for the population mean price (in dollars per 100 pounds) that farmers in this region get for their watermelon crop. Find a 90% confidence interval for the population mean price (in dollars per 100 pounds) that farmers in this region get for their watermelon crop.
(b)
Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.27 for the mean price per 100 pounds of watermelon.
(c)
A farm brings 15 tons of watermelon to market. Find the margin of error for a 90% confidence interval for the population mean cash value (in dollars) of this crop. Find a 90% confidence interval for the population mean cash value of this crop. Hint: 1 ton is 2,000 pounds.

For a 90% confidence level, the Z-score is approximately 1.645. This value is obtained from the standard normal distribution table for a 90% confidence interval (which leaves 5% in each tail).

Calculate the margin of error (ME):

The margin of error formula is:

=

×

ME=Z×

n

σ

Plugging in the values:

=

1.645

×

1.86

37

ME=1.645×

37

1.86

≈

1.645

×

1.86

6.08

ME≈1.645×

6.08

1.86

≈

1.645

×

0.306

ME≈1.645×0.306

≈

0.504

ME≈0.504

Margin of Error (ME) ≈ $0.50

Find the 90% confidence interval:

The confidence interval is given by:

ˉ

±

x

ˉ

±ME

6.88

±

0.50

6.88±0.50

Confidence Interval

=

(

6.38

,

7.38

)

Confidence Interval=(6.38,7.38)

90% Confidence Interval = ($6.38, $7.38)

(b) Sample Size for a 90% Confidence Level

Given Data:

Desired margin of error (

E) = $0.27

Sample standard deviation (

σ) = $1.86

Confidence level = 90%

Steps:

Find the Z-score for a 90% confidence level:

The Z-score is 1.645.

Calculate the required sample size (

n):

The formula for sample size is:

=

(

×

)

2

n=(

E

Z×σ

)

2

Plugging in the values:

=

(

1.645

×

1.86

0.27

)

2

n=(

0.27

1.645×1.86

)

2

=

(

3.0587

0.27

)

2

n=(

0.27

3.0587

)

2

=

(

11.32

)

2

n=(11.32)

2

≈

128.1

n≈128.1

Sample Size

n ≈ 129 (rounding up to ensure sufficient accuracy)

(c) Margin of Error and Confidence Interval for Cash Value

Given Data:

Farm brings 15 tons of watermelon.

1 ton = 2,000 pounds, so 15 tons = 30,000 pounds.

Price per 100 pounds = $6.88

Standard deviation = $1.86 per 100 pounds

Confidence level = 90%

Steps:

Calculate the total cash value of the crop:

Total cash value:

Value

=

30

,

000

100

×

6.88

Value=

100

30,000

×6.88

Value

=

300

×

6.88

Value=300×6.88

Value

=

2

,

064

Value=2,064

Find the margin of error for the total value:

First, find the standard deviation for the total value. Since the price per 100 pounds is given:

Standard deviation for total value

=

30

,

000

100

×

Standard deviation for total value=

100

30,000

×σ

Standard deviation for total value

=

300

×

1.86

Standard deviation for total value=300×1.86

Standard deviation for total value

=

558

Standard deviation for total value=558

Now calculate the margin of error for the total cash value:

=

×

Standard deviation for total value

ME=Z×Standard deviation for total value

=

1.645

×

558

ME=1.645×558

≈

918.51

ME≈918.51

Margin of Error ≈ $918.51

Find the 90% confidence interval for the total cash value:

The confidence interval is given by:

Value

±

Value±ME

2

,

064

±

918.51

2,064±918.51

Confidence Interval

=

(

1

,

145.49

,

2

,

982.51

)

Confidence Interval=(1,145.49,2,982.51)

90% Confidence Interval for the total cash value = ($1,145.49, $2,982.51)