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###### Hypothesis Testing

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1 week ago

Hello everyone

I have worked out the answers for both c and d but I am stuck on the final part for sketching a drawing and comparing the results.

Any help would be appreciated.

c) Your analysis shows that the mean capacitance of a batch of 500 of the capacitors you have selected is 46 μF, with a standard deviation of 4 μF. Assuming the capacitors are normally distributed, determine the number of capacitors likely to have values between 42 μF and 50 μF.

Note: Use the z-table given in Appendix A when answering part c.

d) The Quality Assurance Department is anxious to improve the nominal value of the capacitors to ensure more of them fall within the rated tolerance band. The manufacturing process is adjusted in an attempt to bring about this improvement. Following the adjustment, a further sample of 100 capacitors is taken in order to determine if the adjustment has resulted in an improvement to the nominal value of the capacitors. Analysing this second sample, it is noted that the mean value is now 46.5 μF, with a standard deviation of 2 μF.

By comparing these new results of the capacitor values following the manufacturing adjustment, with the values obtained in part (c) recorded before the adjustment, show whether or not you agree with the hypothesis that the adjustment to the manufacturing process has had a beneficial effect by making the capacitor values closer to the desired manufactured nominal value of 47 μF.

Illustrate your answer with a hand drawn sketch of the normal distribution, with shaded regions representing the results of your analysis, which you can show to the Quality Assurance Manager.

Note: Use the z-table given in Appendix A when answering part d.

I have worked out the answers for both c and d but I am stuck on the final part for sketching a drawing and comparing the results.

Any help would be appreciated.

c) Your analysis shows that the mean capacitance of a batch of 500 of the capacitors you have selected is 46 μF, with a standard deviation of 4 μF. Assuming the capacitors are normally distributed, determine the number of capacitors likely to have values between 42 μF and 50 μF.

Note: Use the z-table given in Appendix A when answering part c.

d) The Quality Assurance Department is anxious to improve the nominal value of the capacitors to ensure more of them fall within the rated tolerance band. The manufacturing process is adjusted in an attempt to bring about this improvement. Following the adjustment, a further sample of 100 capacitors is taken in order to determine if the adjustment has resulted in an improvement to the nominal value of the capacitors. Analysing this second sample, it is noted that the mean value is now 46.5 μF, with a standard deviation of 2 μF.

By comparing these new results of the capacitor values following the manufacturing adjustment, with the values obtained in part (c) recorded before the adjustment, show whether or not you agree with the hypothesis that the adjustment to the manufacturing process has had a beneficial effect by making the capacitor values closer to the desired manufactured nominal value of 47 μF.

Illustrate your answer with a hand drawn sketch of the normal distribution, with shaded regions representing the results of your analysis, which you can show to the Quality Assurance Manager.

Note: Use the z-table given in Appendix A when answering part d.

Original post by thomas0611

Hello everyone

I have worked out the answers for both c and d but I am stuck on the final part for sketching a drawing and comparing the results.

Any help would be appreciated.

c) Your analysis shows that the mean capacitance of a batch of 500 of the capacitors you have selected is 46 μF, with a standard deviation of 4 μF. Assuming the capacitors are normally distributed, determine the number of capacitors likely to have values between 42 μF and 50 μF.

Note: Use the z-table given in Appendix A when answering part c.

d) The Quality Assurance Department is anxious to improve the nominal value of the capacitors to ensure more of them fall within the rated tolerance band. The manufacturing process is adjusted in an attempt to bring about this improvement. Following the adjustment, a further sample of 100 capacitors is taken in order to determine if the adjustment has resulted in an improvement to the nominal value of the capacitors. Analysing this second sample, it is noted that the mean value is now 46.5 μF, with a standard deviation of 2 μF.

By comparing these new results of the capacitor values following the manufacturing adjustment, with the values obtained in part (c) recorded before the adjustment, show whether or not you agree with the hypothesis that the adjustment to the manufacturing process has had a beneficial effect by making the capacitor values closer to the desired manufactured nominal value of 47 μF.

Illustrate your answer with a hand drawn sketch of the normal distribution, with shaded regions representing the results of your analysis, which you can show to the Quality Assurance Manager.

Note: Use the z-table given in Appendix A when answering part d.

I have worked out the answers for both c and d but I am stuck on the final part for sketching a drawing and comparing the results.

Any help would be appreciated.

c) Your analysis shows that the mean capacitance of a batch of 500 of the capacitors you have selected is 46 μF, with a standard deviation of 4 μF. Assuming the capacitors are normally distributed, determine the number of capacitors likely to have values between 42 μF and 50 μF.

Note: Use the z-table given in Appendix A when answering part c.

d) The Quality Assurance Department is anxious to improve the nominal value of the capacitors to ensure more of them fall within the rated tolerance band. The manufacturing process is adjusted in an attempt to bring about this improvement. Following the adjustment, a further sample of 100 capacitors is taken in order to determine if the adjustment has resulted in an improvement to the nominal value of the capacitors. Analysing this second sample, it is noted that the mean value is now 46.5 μF, with a standard deviation of 2 μF.

By comparing these new results of the capacitor values following the manufacturing adjustment, with the values obtained in part (c) recorded before the adjustment, show whether or not you agree with the hypothesis that the adjustment to the manufacturing process has had a beneficial effect by making the capacitor values closer to the desired manufactured nominal value of 47 μF.

Illustrate your answer with a hand drawn sketch of the normal distribution, with shaded regions representing the results of your analysis, which you can show to the Quality Assurance Manager.

Note: Use the z-table given in Appendix A when answering part d.

What are you actually stuck with / what have you tried, so sketching the normal, ...?

Reply 2

1 week ago

Original post by mqb2766

What are you actually stuck with / what have you tried, so sketching the normal, ...?

I don’t know how to draw the normal distribution graph.

Do I need to draw a graph for my answer for c and d?

Original post by thomas0611

I don’t know how to draw the normal distribution graph.

Do I need to draw a graph for my answer for c and d?

Do I need to draw a graph for my answer for c and d?

The normal distribution is centered on the mean value and the width is determined by the standard deviation. If you google it, you could do it in excel, desmos, online tool, ... as well as lots of descriptions. Just try and sketch both distributions and see what you think and post what you have if unsure

Reply 4

6 days ago

I have sketched what I think is the correct answer but I cant upload the picture. Do you know how to post a photo without a url

Thanks

Thanks

Original post by thomas0611

I have sketched what I think is the correct answer but I cant upload the picture. Do you know how to post a photo without a url

Thanks

Thanks

Sorry, I dont.

Reply 6

6 days ago

Original post by mqb2766

Sorry, I dont.

No worries

My answer for the null hypothesis is 1.24%. I have labelled the x axis 1s.ds, 2s.ds etc with my mean 46.5 at the centre

My standard deviation is calculated at 2.5 so I have drawn a line on the right hand side from 2.5s.ds

This has resulted in a very small shaded area to represent 1.24%

Does this sound like i'm on the right track?

Thanks

Original post by thomas0611

No worries

My answer for the null hypothesis is 1.24%. I have labelled the x axis 1s.ds, 2s.ds etc with my mean 46.5 at the centre

My standard deviation is calculated at 2.5 so I have drawn a line on the right hand side from 2.5s.ds

This has resulted in a very small shaded area to represent 1.24%

Does this sound like i'm on the right track?

Thanks

My answer for the null hypothesis is 1.24%. I have labelled the x axis 1s.ds, 2s.ds etc with my mean 46.5 at the centre

My standard deviation is calculated at 2.5 so I have drawn a line on the right hand side from 2.5s.ds

This has resulted in a very small shaded area to represent 1.24%

Does this sound like i'm on the right track?

Thanks

It would help to see the picture / actual working as the stuff sounds like it doesnt match. Why does a difference of 0.5 (mean) with a standard devaition of 2.5 correspond to such a small % value? It would probably help to clearly state the hypothesis test, then state the stats associated with the mean estimate for the new sample, then ...

Reply 8

4 days ago

Original post by mqb2766

It would help to see the picture / actual working as the stuff sounds like it doesnt match. Why does a difference of 0.5 (mean) with a standard devaition of 2.5 correspond to such a small % value? It would probably help to clearly state the hypothesis test, then state the stats associated with the mean estimate for the new sample, then ...

I think I managed to solve the problem I was having. Thanks for your help

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