The distribution of the amount of medication in a particular pill is known to va

Question

The distribution of the amount of medication in a particular pill is known to vary normally with unknown mean and standard deviation 10 mg. A pharmaceutical company that manufactures this pill advertises that they contain 500 mg of a particular medication. The plant manager where the medication is produced believes the machinery is incorrectly calibrated and the amount of medication in the pills is actually more than 500 mg. She takes a simple random sample of 15 pills and has them analyzed for the amount of medication they contain. The amount in the 15 pills, in mg, is: 507 503 493 500 488 514 502 520 489 503 511 495 496 498 504 Perform the appropriate hypothesis test by answering the following questions: Please show all of your work: a. State the null and alternative hypotheses b. Enter the data into the StatCrunch data table. c. Go to STAT/Z STATISTICS/ONE SAMPLE/WITH DATA. Select the variable. Enter 10 for the standard deviation and select NEXT. Enter the correct NULL: MEAN and ALTERNATIVE hypothesis. Click on CALCULATE. d. From the StatCrunch output: i) What is the sample mean? ii) What is the test statistic? iii) What is the p-value? iv) What conclusions can you draw about the machinery? e) suppose instead that the plant manager had taken an SRS of size n = 50 pills. Repeat the previous problem using the same sample mean (501.53) and same population standard deviation (10) but assume that the sample mean was found from an SRS of size n = 50. Go to STAT/Z STATISTICS/ONE SAMPLE/WITH SUMMARY. Enter 501.53 for the mean, 10 for the standard deviation, and 50 for the sample size. Select NEXT. Enter the correct NULL: MEAN and ALTERNATIVE hypothesis. Click on CALCULATE. What conclusions can you draw about the machinery? f) Repeat the previous problem with n = 200. What conclusions do you now draw about the machinery? How does changing the sample size change the hypothesis test results?

Explanation / Answer

The distribution of the amount of medication in a particular pill is known to vary normally with unknown mean and standard deviation 10 mg. A pharmaceutical company that manufactures this pill advertises that they contain 500 mg of a particular medication. The plant manager where the medication is produced believes the machinery is incorrectly calibrated and the amount of medication in the pills is actually more than 500 mg. She takes a simple random sample of 15 pills and has them analyzed for the amount of medication they contain. The amount in the 15 pills, in mg, is: 507 503 493 500 488 514 502 520 489 503 511 495 496 498 504

Here we have to test the hypothesis that,

H0 : mu = 500 Vs H1 : mu > 500

where mu is population mean

Assume alpha= level of significance = 0.05

Here sample data is given and sample size is too small so we use one sample t-test.

————— 22-11-2017 09:23:33 ————————————————————

Welcome to Minitab, press F1 for help.

One-Sample T: C1

Test of mu = 500 vs mu > 500

Variable N Mean StDev SE Mean
C1 15 501.53 8.98 2.32

Variable 95.0% Lower Bound T P
C1 497.45 0.66 0.259

Test statistic = 0.66

P-value = 0.259

P-value > alpha

Accept H0 at 5% level of significance.

Conclusion : There is sufficient evidence to say that the population mean may be 500.

e) suppose instead that the plant manager had taken an SRS of size n = 50 pills. Repeat the previous problem using the same sample mean (501.53) and same population standard deviation (10) but assume that the sample mean was found from an SRS of size n = 50.

n = 50

Xbar = 501.53

sigma = 10

Now here sample size is >30 also population standard deviation is known so we use one sample z-test.

Test statistic = 0.94

P-value = 0.1735

P-value > alpha

Accept H0 at 5% level of significance.

Conclusion :  There is sufficient evidence to say that the population mean may be 500.

f) Repeat the previous problem with n = 200. What conclusions do you now draw about the machinery? How does changing the sample size change the hypothesis test results?

Now we have to do same procedure with n = 200

Test statistic = 1.88

P-value = 0.02999

P-value < alpha

Reject H0 at 5% level of significance.

Conclusion : There is sufficient evidence to say that population mean is greator than 500.

Now we can see that as samplee size increases we get significant result.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.