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A company manufacturing some medicine tablets with the Average weight of 500 mg and population variance of 36 mg. A quality officer in a company wants to investigate whether the weight of Paracetamol tablets is differed from 500 mg. To investigate He taken 50 tablets at the time of manufacturing and find that the average weight of 50 tablets is 504.5 mg.  Make the hypothesis test that the average weight of tablets is differ from or not at 95% confidence level.  INSTRUCTIONS: You can perform this exercise in Minitab? the exercise is already solved! But I just want to see how to do it in Minitab, please!

A company manufacturing some medicine tablets with the Average weight of 500 mg and population variance of 36 mg. A quality officer in a company wants to investigate whether the weight of Paracetamol tablets is differed from 500 mg. To investigate He taken 50 tablets at the time of manufacturing and find that the average weight of 50 tablets is 504.5 mg.  Make the hypothesis test that the average weight of tablets is differ from or not at 95% confidence level.  INSTRUCTIONS: You can perform this exercise in Minitab? the exercise is already solved! But I just want to see how to do it in Minitab, please!

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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A company manufacturing some medicine tablets with the Average weight of 500 mg and population variance of 36 mg. A quality officer in a company wants to investigate whether the weight of Paracetamol tablets is differed from 500 mg. To investigate He taken 50 tablets at the time of manufacturing and find that the average weight of 50 tablets is 504.5 mg. 

Make the hypothesis test that the average weight of tablets is differ from or not at 95% confidence level. 


INSTRUCTIONS:
You can perform this exercise in Minitab? the exercise is already solved!

But I just want to see how to do it in Minitab, please!

Note:

the exercise is already solved! as seen in the image I attached, I just want to see it solved in minitab!

(with pictures, or sreenshots) however, but I want to see the procedure :)

Step 1- Hypotheses -: Null hypothesis H,: u = 500 (or) The population average weight of tablets is equal to 500 mg Alternative hypothesis H;: u# 500 (or) The population average weight of tablets is not equal to 500 mg Step 2- Calculation of z test statistic: Given values are, x bar = sample mean = 504.5 mg H = population mean = 500 mg o = Population standard deviation =6 [ because population variance o = 36] %3D n= sample size = 50 Substitute the above values in z test statistic formula as given = z vn 504.5-500 4.5 6/7.071 4.5 0.85 5.30 Thus, the z test statistic is 5.30
Transcribed Image Text:Step 1- Hypotheses -: Null hypothesis H,: u = 500 (or) The population average weight of tablets is equal to 500 mg Alternative hypothesis H;: u# 500 (or) The population average weight of tablets is not equal to 500 mg Step 2- Calculation of z test statistic: Given values are, x bar = sample mean = 504.5 mg H = population mean = 500 mg o = Population standard deviation =6 [ because population variance o = 36] %3D n= sample size = 50 Substitute the above values in z test statistic formula as given = z vn 504.5-500 4.5 6/7.071 4.5 0.85 5.30 Thus, the z test statistic is 5.30
Step 3:z-table value: Given Significance level is a = 0.05. Now, Za/z is the z critical value at 95% confidence level is +1.96 (refer from the standard normal z table values i.e., za/2- Z0.05/2 =+1.96 step 4 - Rejection rule: we rejecting null hypothesis H, if z test statistic value outside the range of z critical values. otherwise accept null hypothesis H, at 95% confidence level. Step 5 - Conclusion: Since, the z test statistic (5.30) is outside the z critical values of (-1.96 to +1.96). Thus, we have sufficient evidence to reject the null hypothesis H, at 95% confidence level. Hence, null hypothesis H, is rejected. So, the average weight of tablets is not equal to 500 mg
Transcribed Image Text:Step 3:z-table value: Given Significance level is a = 0.05. Now, Za/z is the z critical value at 95% confidence level is +1.96 (refer from the standard normal z table values i.e., za/2- Z0.05/2 =+1.96 step 4 - Rejection rule: we rejecting null hypothesis H, if z test statistic value outside the range of z critical values. otherwise accept null hypothesis H, at 95% confidence level. Step 5 - Conclusion: Since, the z test statistic (5.30) is outside the z critical values of (-1.96 to +1.96). Thus, we have sufficient evidence to reject the null hypothesis H, at 95% confidence level. Hence, null hypothesis H, is rejected. So, the average weight of tablets is not equal to 500 mg
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