Question 1Sample meanStandard deviation of populationSamplevalue would be used because sample size...

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Question 1Sample meanStandard deviation of populationSamplevalue would be used because sample size is higher than 30 and population’s standard deviation is knownvalue for 90% confidence interval = 1.6490% confidence interval =Upper limit of 90% confidence interval = = 5.5 daysLower limit of 90% confidence interval = = 4.9 daysMarginal of error = daysIt can be stated with 90% confidence that the mean days of patient stays would lie between 4.9 days and 5.5 days.			Document Preview:				Mike Wynn  Homework 4-5-17  Question 1  Sample mean  x=5.2 days  Standard deviation of population s=1.9 days  Sample n=100  z value would be used because sample size is higher than 30 and population’s standard deviation is known  z value for 90% confidence interval  = 1.64  90% confidence interval =  x±z.(s/n) Upper limit of 90% confidence interval = 5.2+1.64*(1.9/100) = 5.5 days  Lower limit of 90% confidence interval = 5.2-1.64*(1.9/100) = 4.9 days  Marginal of error = Upper limit-mean=5.5-5.2=0.3 days  It can be stated with 90% confidence that the mean days of patient stays would lie between 4.9 days and 5.5 days.   Question 2  Sample mean  x=$200  Standard deviation of sample s=$30  Sample n=40  Sample’s standard deviation is given and thus, t value would be used. Degree of freedom = n-1=40-1=39   t value = 2.02 95% confidence interval = x ±t.(s/n) Upper limit of 95% confidence interval = 200+2.02×3040=$209.5  Lower limit of 95% confidence interval = 200-2.02×3040=$190.4   Marginal of error = Upper limit-mean=209.5-200=$9.58 It can be sated with 95% confidence that the mean deductible amount of employee health insurance plan would lie between$209.5  and $190.4     Question3 xi Weight of bag of salted peanuts (in ounces) 15.417.216.016.315.814.917.216.516.816.417.017.316.116.315.916.5  xixi- x(xi- x)215.4-0.950.902517.20.850.722516-0.350.122516.3-0.050.002515.8-0.550.302514.9-1.452.102517.20.850.722516.50.150.022516.80.450.202516.40.050.0025170.650.422517.30.950.902516.1-0.250.062516.3-0.050.002515.9-0.450.202516.50.150.0225Total  = (16.35) Total  = (6.72)  Number of samples  n=16  Mean of the sample x=xin=16.35  x= 16.35 ounces   Standard deviation of sample s= 1n-1 (xi- x)2 = 116-1 *6.72   s=0.67 ounces  Degree of freedom =  n-1=16-1=15 As the sample size is lower than 30 and sample standard deviation is computed and thus, t value would be used. t value= 2.94  99%...

David · Accepted Answer

1) Hypothesis testing:
When we know population standard deviation and also sample size >30 
We can use Z critical value to find confidence interval. 
For 90% two tailed z critical value = 1.64 (From std normal table) 
Sample mean = 5.2 and sigma = 1.9 
Hence margin of error =  
 (    )
√ )
           
For 90% confidence interval we can have the interval as 
Mean            
= (5.2-0.31, 5.2+0.31) 
= (4.89, 5.51) 
Interpretation of confidence interval: It can be stated with 90% confidence that the 
mean days of patient stays would lie between 4.9 days and 5.5 days.  
2) Here sample size is greater than 40, but population std deviation is not known.  
Only sample std deviation s is given. 
Hence critical value of t with degree of freedom n-1 = 39 should be used. 
T value = 2.02
Margin of error =    (
 
√ )
  (    )  (
  
√  
)
Hence upper limit of confidence interval = Mean + Margin of error 
= 200+9.58 = 209.58 
Lower limit = 200-9.58 = 191.

Question 1 Sample mean Standard deviation of population Sample value would be used because sample size is higher than 30 and population’s standard deviation is known value for 90% confidence interval...

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