Question 567814
I would like assistance with the following 2 math problems. I did Exercise 32. May I please have you check it for errors along with assisting me with finding the p-value. (Please share with me formula, how you used the z-test, or if done in Excel a "prints screen" shot and direction please as to how to replicate the solution. Lastly, I need help with the second problem, Exercise 46 Thank you
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Exercise Question 32: Dole Pineapple, Inc. is concerned that the 16-ounce can of sliced pineapple is being overfilled. Assume the standard deviation of the process is .03 ounces. The quality control department took a random sample of 50 cans and found that the arithmetic mean weight was 16.05 ounces. At the 5 percent level of significance, can we conclude that the mean weight is greater than 16 ounces? Determine the P-value.
Note: Equality must ALWAYS be part of Ho. 
Ho: &#956;<=16
Ha: &#956;>16 (claim)
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&#945; = 0.05
critical value z= 1.645
Test statistics: Z= (xbar -µ)/ (&#963;/&#8730;n)
(16.05-16.0)/ (0.03/&#8730;50)
(0.05)/ (0.03/7.071)
0.05/0.0042 = 11.90 
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p-value = P(z > 11.90) = normalcdf(11.90,100) = 6.09x10^-33
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Determine the P-value (please show applicable formula and all work – how was answer derived. If using Excel please “PrintScreen” and attach. Thanks) 
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Conclusion: Since the p-value is less than 5%, reject Ho.
The test results support the claim.
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Grand Strand Family Medical Center is specifically set up to treat minor medical emergencies for visitors to the Myrtle Beach area. There are two facilities, one in the Little River Area and the other in Murrells Inlet. The Quality Assurance Department wishes to compare the mean waiting time for patients at the two locations. Samples of the waiting times reported in minutes follow: 
Location Waiting Time 
Little River 31.73 28.77 29.53 22.08 29.47 18.60 32.94 25.18 29.82 26.49
Murrells Inlet 22.93 23.92 26.92 27.20 26.44 25.62 30.61 29.44 23.09 23.10 26.69 22.31
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Little River mean and std: mean =27.46  ; std = 4.44 
Murrell's Inlet mean and std: mean =25.69   ; std = 2.68
 
Assume the population standard deviations are not the same. 
Ho: u1-u2 = 0
Ha: u1-u2 # 0 (claim)
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t(x1bar-x2bar) = (27.46-25.69)/sqrt[(4.44^2/10)+(2.68^2/12)] = 1.1041
df = n1+n2-2 = 20
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p-value = 2*P(t > 1.1041 when df = 20) = 2*tcdf(1.1041,100,20) = 0.2826

At the .05 significance level, is there a difference in the mean waiting time?
Since the p-value is greater than 5%, fail to reject Ho.
The test results do not support the claim.
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Cheers,
Stan H.
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