Question

Jerry has 30 bags of candy. Each bag states that the net weight of the bag...

Jerry has 30 bags of candy. Each bag states that the net weight of the bag is .0469 pounds. Jerry believes that the bags of candy will not all weigh .0469 pounds, so he weighs each bag and writes down the results below.

1) .0466 2) .0471 3).0468 4).0465 5) .0461 6) .0454 7) .0470 8) .0463 9) .0469 10) .0473 11).0462 12) .0468 13) .0469 14).0471 15) .0470 16) .0468 17) .0481 18) .064 19) .0475 20) .0469 21) .0467 22) .0460 23) .0473 24) .0466 25) .0470 26) .0471 27) .0467 28) .0472 29) .0469 30) .0468

1. What is the Null and Alternate hypothesis statement? Indicate which is the claim. 2. State the significance level. Justify why this level is appropriate. 3. Make a box plot for the data and discuss whether outliers exist in it. 4. Perform the appropraite hypothesis test. List the calcutator function and the input used. 5. Report the test statistic and p-value. 6. State the decision the test indicates. 7. Write the conclusion in an English sentence in terms of the problem stated as the research question. 8. Explain how the results of a t-test provide a statistical answer to the research question. Is this proof that your result is true?

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Answer #1

Solution:

a. Null Hypothesis (Ho): µ = 0.0469

Alternative Hypothesis (Ha): µ ? 0.0469 (claim)

b. Level of significance,a = 0.05.  It is indeed the default value in most statistical software applications.

c.

BoxPlot 0.06 0.065 0.07 0.055 #1 0.04 0.045 0.05

From the box plot, we can observe that there lower and higher outlier.

4. Test Statistics

t = (X-bar - µ)/ (s/?n)

t = (0.04739 - 0.0469)/ (0.003177/?30)

t = 0.84

for X-bar, we use =AVERAGE and for s, we use = STDEV.S

5. T-statistic = 0.85, degrees of freedom, df = n - 1 = 30 - 1 = 29

Using t-tables, the p-value is

P-value = 0.4083

6. Since p-value is greater than 0.05 level of significance, we fail to reject the null hypothesis.

7. Hence, we cannot conclude that the net weight of the bag is .0469 pounds.

8. T-test helps in rejecting or fail to reject the null hypothesis which leads to the statistical answer to the research question. Hence, this proofs that the result is true.

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