how to caculate P value? An aspirin man acturer fils bottles by weight rather th

Question

how to caculate P value?

An aspirin man acturer fils bottles by weight rather than b count. Since each bottle should contain 100 tablet the averaoe weight per tablet should be grains. Each of 100 tablets ta en rom a vcr large lot is weighed result ng in sample avcragc wcight Per tablet of 4.a6 grains and a sample standard deviation of 0.36 qrain, Does this intormation provide strong evidence for concluding that the company is not tilling its bottles as advertised? Test the appropriate hypotheses using-0.01 by tirst computing the P value and then comparing it to the specified significancc levcl. State the appropriate hypotheses, Ha: = 5 Calculate tha tast statisc valua and datarmine tha p-value. (Round yaur 7 to twn dacimal places. Round your P-valus to three decimal places.) z=1-3 89 p-value- Stata the conclusion in the Do not reject tha null hypath sls. Thana 1 sufficient avid nca to con uda that me company I, not filling its hottias as advertise. ·I Reject the null hypothesis. There is sufficient evidence to conclude that the company is not filling its bottles as advertised. Raject the null hypathesis. Thera is not sufficiant avidence to concluda that the company is not filling its bottlas as advertisad. D not reject tha null hypothesis. There is nat sufficiant evidence to conclud that the company it not filling its bottles as advartisea

Explanation / Answer

Here It is given that the aspirin manufacturer is filling bottles based upon weight and each bottle should have 100 tablets of 5gm each.

Now the samples of n=100 tablets is taken anbd found that avg weight/tablet is 4.86gm with std deviation= 0.36gm. we have to test the hypotheisi at alpha=0.01 i.e 99% significance

So here we are testing for the following case:

Case : The avg weight is less than 5gm

So the hypothesis is

Ho: U=5

Ha: U<5 [Which is the case given above]

Here we have to take Z test since we know the population sd

Z-stat=(X'-U)/(S.D. * Sqrt(n))=(4.86-5)/(0.36*sqrt(100))= -0.03888889

From the z - table we know that z value for p = 0.01 is -2.326348 (Since it's a one tail test)

While the caluclated z value is -0.0389 which is more than the z value provided through the table i.e. -2.326348 and so here we can say that we are accepting the null hypothesis which states that the supplier provides drugs of 5gms only with p value of 0.01 i.e. nothing but at significance level of 99%

Alfa level is nothing but the P value which is our refernce

Hope this has helped you in understanding the proble. Pls. upvote the ans if it has really helped you. Good Luck!!

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