Question

As the quality control manager at a plant that produces cereal, you would like to ensure that the average amount of cereal being put in each box is 455g. A random sample is contained below (this sample is also contained in the DATA 2 tab of the downloaded Excel file). At a 5% significance level, can you conclude that the average fill level is different than 455g?

Cereal Box Fill Levels
451.48	453.62	452.03	455.14	457.1	455.61	458.61	458.96	452.53	452.54	455.43	452.32	455.12
457.6	455.6	457.44	458.39	457.57	456.79	457.06	456.23	457.05	453.04	458.26	455.75	453.37
456.89	458.99	457.79	457.68	456.13	457.85	458.65	458.98	457.98	457.46	455.09	457.72	458.69
456.22	455.58	451.83	453.19	458.58	455.72	456.69	457.85	453.13	458.83	457.25	452.78	454.92
453.69	451.17	455.72	455.38	458.16	457.24	453.28	451	454.92	457.65	452.88	456.1	452.62

a) Properly state the hypotheses:(1 mark)

b) Calculate the p-value (SHOW ALL WORK):

N ORDER TO SHOW YOUR WORK FOR PART B...CREATE A TABLE THAT MIMICS THE EXCEL FUNCTION YOU ARE USING...YOUR ANSWER SHOULD INCLUDE THE FOLLOWING:

1) FUNCTION NAME:

2) FIELD 1 ENTRY:

3) FIELD 2 ENTRY:

4) FIELD 3 ENTRY:

5) FIELD 4 ENTRY:

6) YOUR FINAL ANSWER:

IF THERE ARE NOT 4 "ARGUMENTS" FOR THE EXCEL FUNCTION, INCLUDE ONLY THE NUMBER OF FIELDS THAT ARE APPLICABLE.

YOUR FIELD ENTRIES MUST BE NUMERICAL VALUES...NOT CELL REFERENCES

c) Conclusion: (Must be mathematically justified) (1 mark)

Answer 1

a. Ho:  the average amount of cereal being put in each box is 455g

H1:  the average amount of cereal being put in each box is significantly different from 455g

alpha = , mu =  455g

451.48
453.62
452.03
455.14
457.1
455.61
458.61
458.96
452.53
452.54
455.43
452.32
455.12
457.6
455.6
457.44
458.39
457.57
456.79
457.06
456.23
457.05
453.04
458.26
455.75
453.37
456.89
458.99
457.79
457.68
456.13
457.85
458.65
458.98
457.98
457.46
455.09
457.72
458.69
456.22
455.58
451.83
453.19
458.58
455.72
456.69
457.85
453.13
458.83
457.25
452.78
454.92
453.69
451.17
455.72
455.38
458.16
457.24
453.28
451
454.92
457.65
452.88

count	63	p-value	0.00523
mean	455.8444444	t_test_value=	2.89473066
sample stdev	2.315438214	t_crit_value	1.998971517

As sample size is 63, we will use t-test as we are not given population standard deviation.

b. For t_test_value = (455.844-455)*sqrt(63)/2.31

for t_crit_value = t.inv.2t(0.05,62)

for p-value = t.dist.2t(t_test_value,62)

As size increases, p-value decreases and test becomes more significant

p-value	0.005232995	0.005612	0.006119	0.007011	0.008959	0.015977	0.033999	0.044352	0.062771	0.101495	0.211753
n	62	50	40	30	20	10	5	4	3	2	1

c. As p-value < 0.05 and t_value > 1.99 , we reject Ho or can say that the average amount of cereal being put in each box is significantly different from 455g.

Please rate my answer and comment for doubt