Question

In: Statistics and Probability

What price do farmers get for their watermelon crops? In the third week of July, a...

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     %

Solutions

Expert Solution

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 )


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