Question

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 41 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.90 per 100 pounds.

(a) A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop. What is the margin of error? Hint: 1 ton is 2000 pounds. (Round your answers to two decimal places.)

lower limit	$
upper limit	$
margin of error	$

Over the past several months, an adult patient has been treated for tetany (severe muscle spasms). This condition is associated with an average total calcium level below 6 mg/dl. Recently, the patient's total calcium tests gave the following readings (in mg/dl). Assume that the population of x values has an approximately normal distribution.

9.5

8.8

10.5

9.1

9.4

9.8

10.0

9.9

11.2

12.1

(b) Find a 99.9% confidence interval for the population mean of total calcium in this patient's blood. (Round your answer to two decimal places.)

lower limit	mg/dl
upper limit	mg/dl

What percentage of hospitals provide at least some charity care? Based on a random sample of hospital reports from eastern states, the following information is obtained (units in percentage of hospitals providing at least some charity care):

57.1

56

53.1

66

59.0

64.7

70.1

64.7

53.5

78.2

(c) Find a 90% confidence interval for the population average μ of the percentage of hospitals providing at least some charity care. (Round your answers to one decimal place.)

lower limit	%
upper limit	%

Answer 1

a)

population std dev ,    σ =    1.9000
Sample Size ,   n =    41
Sample Mean,    x̅ =   6.8800
Level of Significance ,    α =    0.1          
'   '   '          
z value=   z α/2=   1.6449   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   1.900   / √   41   =   0.2967
margin of error, E=Z*SE =   1.6449   *   0.297   =   0.488
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    6.88   -   0.488   =   6.3919
Interval Upper Limit = x̅ + E =    6.88   -   0.488   =   7.3681

now, 1 ton=2000 pounds

15 ton = 30000 pounds=300 hundred pounds

so, margin of error = 0.488*300=146.42

lower limit=6.3919*300=1917.58

upper limit=7.3681*300=2210.42

b)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   1.0001
Sample Size ,   n =    10
Sample Mean,    x̅ = ΣX/n =    10.0300

Level of Significance ,    α =    0.001          
degree of freedom=   DF=n-1=   9          
't value='   tα/2=   4.7809   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   1.000   / √   10   =   0.3162
margin of error , E=t*SE =   4.7809   *   0.316   =   1.512
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    10.03   -   1.512   =   8.5181
Interval Upper Limit = x̅ + E =    10.03   -   1.512   =   11.5419
99.9%   confidence interval is (   8.52   < µ <   11.54   )

c)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   8.0175
Sample Size ,   n =    10
Sample Mean,    x̅ = ΣX/n =    62.2400

Level of Significance ,    α =    0.1          
degree of freedom=   DF=n-1=   9          
't value='   tα/2=   1.8331   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   8.018   / √   10   =   2.5354
margin of error , E=t*SE =   1.8331   *   2.535   =   4.648
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    62.24   -   4.648   =   57.5924
Interval Upper Limit = x̅ + E =    62.24   -   4.648   =   66.8876
90%   confidence interval is (   57.6 < µ <   66.9 )