Question

In the automated packaging process, equipment is set to fill boxes with a mean weight of...

In the automated packaging process, equipment is set to fill boxes with a mean weight of 574 grams. This is the standard for the population. Population standard deviation is not known.

1. Calculate a 95% confidence interval for the mean weight of the cereal boxes, using the sample data given in the Excel spreadsheet labeled Cereal Weights.

2. Based on the confidence interval for the mean weight of cereal boxes, does this sample of 400 boxes provide evidence that the packaging process is meeting the company standard? Explain.

3. The company engineer decided to make adjustments to the filling equipment to correct the issue. A random sample of 30 boxes was selected and weighed after the adjustment. The data for this sample can be found in the Excel spreadsheet labeled Weights after Adjustment. Find the descriptive statistics and the 95% confidence interval for the mean weight of packages after the equipment adjustment. Using the descriptive statistics and the confidence interval, does it appear that the equipment adjustment had any effect on the weight of the packages? Explain.

4. Using the data from the Weights after Adjustment, conduct a hypothesis test, using this sample data, to determine whether the mean weight of cereal boxes is meeting the company standard mean weight 574 grams after the equipment adjustment, with the level of significance of .05. Consider carefully whether a one-tail test or a two-tail test would be more useful to management. Use the company standard mean weight as the population mean; population standard deviation is not known.

Null hypothesis: ____________________________________________________

Alternative hypothesis: ______________________________________________

Critical value: ______________________________________________________

Test statistic: ______________________________________________________ (Must show calculation of test statistic.)

p-value: ___________________________________________________________

Decision: __________________________________________________________

What conclusions can you draw about the population mean weight based on the hypothesis test? Was the equipment adjustment enough to meet the company standard?

589.3672
589.3955
558.3806
579.1045
579.1328
558.409
558.4373
561.6692
585.1714
591.8053
561.7259
578.9344
578.9911
588.6017
588.7718
566.5738
593.5346
593.563
549.5638
595.9444
587.4677
566.6588
571.3082
587.4963
571.6484
572.9526
592.1171
592.3156
572.9525
572.9809
573.0376
579.0478
585.2281
573.4912
573.5479
573.6046
573.8597
573.8882
581.0039
571.5917
581.0323
573.8883
573.8881
590.4161
573.9731
574.3984
575.6174
576.3829
563.7388
588.82
576.431
576.4594
576.998
577.1114
577.0547
577.2248
577.2815
577.4516
577.48
577.5083
587.2325
577.5085
577.6215
591.8818
577.6222
579.0961
586.3535
584.6809
584.7092
577.622
577.9619
585.1629
591.9101
580.6835
561.8024
578.0186
587.3458
578.2738
591.8532
578.2738
578.8976
600.4151
579.0392
581.9876
564.5524
584.9927
591.9103
579.1243
590.2376
590.2942
579.3794
571.3564
587.6293
579.0959
579.3227
577.5366
579.3511
585.531
579.4078
582.4653
584.8467
579.687
577.5324
580.7076
583.7127
577.5891
579.6868
579.772
577.3339
579.8287
587.9368
588.1069
579.8572
579.9988
580.1973
566.646
580.3107
580.3674
580.6792
580.7643
574.3288
590.2331
574.3572
590.2899
586.0659
599.4753
573.4216
580.9911
581.1895
581.2179
581.5297
588.3621
588.4188
581.5864
581.6148
578.2411
566.5609
593.6352
573.3933
581.6149
581.8699
573.0247
581.9266
585.5273
581.9267
579.1483
581.9551
585.0167
582.1251
582.0071
582.1488
582.1772
588.159
588.1874
582.2055
582.3473
582.3756
582.404
582.4323
583.056
583.056
583.7081
587.2235
583.7648
583.7932
586.1462
583.8215
583.8498
583.9349
583.9632
583.9636
583.9631
584.3034
584.3601
590.3136
590.342
590.3703
584.3885
584.5302
584.5586
584.5869
584.8421
584.5869
583.056
584.6436
584.7287
592.4115
584.757
581.6104
584.7854
584.8137
584.8422
585.0122
585.0405
585.0407
585.0687
587.7621
585.0688
585.2107
585.2391
585.2675
585.2677
585.2674
585.3241
593.8292
585.5793
585.6076
576.5923
577.1026
585.636
591.1926
591.9013
582.0354
585.6358
581.1567
588.9813
581.185
585.6359
585.6361
585.6643
574.0408
574.0692
585.6927
585.7494
585.9478
594.8497
586.0048
586.0045
604.2903
586.0329
586.1179
586.1746
573.5589
586.203
577.1593
586.2029
595.1332
586.2031
586.3731
599.5842
600.4914
586.3731
582.0639
586.4014
586.4298
586.4582
574.0978
574.4944
586.7983
586.6667
586.6953
586.7236
586.752
587.1772
580.6851
587.2059
591.8833
587.2055
587.3757
587.404
587.4324
587.4891
587.4892
592.7906
583.7185
587.5741
583.8603
587.6592
587.6875
578.3037
578.8706
587.6875
587.7159
590.5792
587.7726
588.0277
588.1413
588.1978
581.9892
582.1309
588.2262
583.7752
588.453
580.2598
580.2882
588.5099
588.5097
588.538
588.5664
588.7932
588.9916
574.3914
575.5537
588.9919
603.7336
589.0483
589.0486
589.2468
589.4615
589.4617
589.4898
566.8382
571.4309
589.5182
589.5749
590.2836
592.4949
590.3403
590.3687
590.7089
594.0542
579.0287
590.7111
590.7656
590.7939
591.1058
591.1625
603.7215
603.8066
591.1625
591.9563
591.9847
582.0905
591.9848
592.013
576.0803
576.1086
576.5055
592.0415
592.0413
592.1264
592.1831
575.7684
575.8251
576.0236
592.1832
592.1833
592.2114
592.2398
592.4666
592.495
580.8149
592.8068
592.8069
592.8351
592.8635
592.8636
603.8068
603.8349
592.7421
592.7705
592.7707
592.7988
580.6936
592.8272
593.5359
593.5926
583.0181
583.0464
593.621
593.6212
593.6496
593.6499
593.6493
593.6777
593.706
593.7062
593.8478
593.8761
593.9045
593.9612
594.6132
579.8429
580.098
580.1264
594.6418
594.6416
594.6983
594.9818
580.2681
594.9818
595.0385
595.0668
597.0797
571.338
571.5363
597.1364
597.1931
579.0779
599.4044
599.4327
600.425
600.4533
580.7217
603.4869
603.4868
581.147
603.5151

566.6034

603.7136

Homework Answers

Answer #1

I used R software to solve this question.

R codes:

weight=scan('clipboard')
Read 400 items

t.test(weight,mu=574)

One Sample t-test

data: weight
t = 25.714, df = 399, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 574
95 percent confidence interval:
583.8503 585.4812
sample estimates:
mean of x
584.6658

Que.1

95% confidence interval for mean weight of cereal box is,

Lower bound = 583.8503

Upper bound =  585.4812

Que.2

Since 95% confidence interval does not contain value 574, hence we conclude that packaging process does not meet the company's standard.

Please provide data 'Weights after adjustment' for solving question no.3 and 4.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The makers of Mini-Oats cereal have an automated packaging machine that is set to fill boxes...
The makers of Mini-Oats cereal have an automated packaging machine that is set to fill boxes with 24.5 ounces of cereal (as labeled on the box). At various times in the packaging process, we select a random sample of 100 boxes to see if the machine is (on average) filling the boxes as labeled. On Tuesday morning, at 7:45 a.m., a random sample of 100 boxes produced an average amount of 23.5 ounces. Which of the following is an appropriate...
Construct the indicated confidence interval for the population mean of each data set. If it is...
Construct the indicated confidence interval for the population mean of each data set. If it is possible to construct a confidence interval, justify the distribution you used. If it is not possible, explain why. (a) In a random sample of 40 patients, the mean waiting time at a dentist’s office was 20 minutes and the standard deviation was 7.5 minutes. Construct a 95% confidence interval for the population mean. (b) In a random sample of 20 people, the mean tip...
The packaging process in a breakfast cereal company has been adjusted so that an average of...
The packaging process in a breakfast cereal company has been adjusted so that an average of μ = 13.0 oz of cereal is placed in each package. The standard deviation of the actual net weight is σ = 0.1 oz and the distribution of weights is known to follow the normal probability distribution. 1. what proportion of cereal packages contain more than 12.9 oz? 2. Suppose 25 cereal boxes are chosen at random, what is the mean weight of all...
Construct the indicated confidence interval for the population mean of each data set. If it is...
Construct the indicated confidence interval for the population mean of each data set. If it is possible to construct a confidence interval, justify the distribution you used. If it is not possible, explain why. ***Please provide formulas used, step by step process, and hand write....thank you so much! In a random sample of 40 patients, the mean waiting time at a dentist’s office was 20 minutes and the standard deviation was 7.5 minutes. Construct a 95% confidence interval for the...
HT Mean: A Packaging Company produces boxes out of cardboard and has a specified weight of...
HT Mean: A Packaging Company produces boxes out of cardboard and has a specified weight of 35 oz. It is known that the weight of a box is normally distributed with standard deviation 1.3 oz. A random sample of 36 boxes yielded a sample mean of 35.5 oz. At 5% level of significance, test the claim that the mean weight of a box is 35 oz or is there significant evidence that the mean weight is greater than 35 oz....
HT Mean: A Packaging Company produces boxes out of cardboard and has a specified weight of...
HT Mean: A Packaging Company produces boxes out of cardboard and has a specified weight of 35 oz. It is known that the weight of a box is normally distributed with standard deviation 1.3 oz. A random sample of 36 boxes yielded a sample mean of 35.5 oz. At 5% level of significance, test the claim that the mean weight of a box is 35 oz or is there significant evidence that the mean weight is greater than 35 oz....
Granola Crunch cereal is packaged in 1 pound boxes. Susan Torres, a quality control analyst who...
Granola Crunch cereal is packaged in 1 pound boxes. Susan Torres, a quality control analyst who works for the manufacturer of the cereal, wants to check if the packaging process is working improperly: boxes are being over-filled or under-filled. A sample of 40 cereal boxes of Granola Crunch yields a mean weight of 1.02 pounds of cereal per box. Assume that the weight is normally distributed with a population standard deviation of 0.04 pound. Conduct a hypothesis test at the...
a. What is the minimum sample size required to estimate the overall mean weight of boxes...
a. What is the minimum sample size required to estimate the overall mean weight of boxes of Captain Crisp cereal to within 0.02 ounces with 95% confidence, if the overall (population) standard deviation of the weights is 0.23 ounces? Enter an integer below. You must round up. b. What is the minimum sample size required to estimate the overall proportion of boxes that have less than 16 ounces of cereal to within 5% with 95% confidence, if no guess as...
A cereal producer makes cereal boxes with a stated weight of 759 grams. From a large...
A cereal producer makes cereal boxes with a stated weight of 759 grams. From a large lot of these boxes, a random sample of 7 boxes yielded an average of 795.3 grams and standard deviation of 17.8 grams. Construct a 99% confidence interval for µ, the mean weight of boxes in this lots. Based on your confidence interval, what business suggestion will you make to the producer?
A cereal producer makes cereal boxes with a stated weight of 759 grams. From a large...
A cereal producer makes cereal boxes with a stated weight of 759 grams. From a large lot of these boxes, a random sample of 7 boxes yielded an average of 795.3 grams and standard deviation of 17.8 grams. Construct a 99% confidence interval for µ, the mean weight of boxes in this lots. Based on your confidence interval, what business suggestion will you make to the producer?