Assignment:
Be sure to state clear hypotheses, test statistic values, critical value or p-value, decision (reject/fail to reject), and conclusion in English. When doing calculations for the color proportions, keep at least 4-6 decimal places sample proportions, otherwise you will encounter large rounding errors.
Master foods USA states that their color blends were selected by conducting consumer preference tests, which indicated the assortment of colors that pleased the greatest number of people and created the most attractive overall effect. On average, they claim the following percentages of colors for M&Ms® milk chocolate candies: 24% blue, 20% orange, 16% green, 14% yellow, 13% red and 13% brown.
1 Test their claim that the true proportion of blue M&Ms® candies is 0.24 at the 0.05 significance level.
Z Test of Hypothesis for the Proportion
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|
|
Data
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Null Hypothesis p=
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0.24
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Level of Significance
|
0.05
|
Number of Successes
|
1967
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Sample Size
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9194
|
|
|
Intermediate Calculations
|
Sample Proportion
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0.213943876
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Standard Error
|
0.004454102
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Z Test Statistic
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-5.849916104
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|
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Two-Tail Test
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|
Lower Critical Value
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-1.959963985
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Upper Critical value
|
1.959963985
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p-Value
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4.91821E-09
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Reject the null hypothesis
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|
2 Test their claim that the true proportion of orange M&Ms® candies is 0.20 at the 0.05 significance level.
Details
Z Test of Hypothesis for the Proportion
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|
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Data
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Null Hypothesis p=
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0.2
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Level of Significance
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0.05
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Number of Successes
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1973
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Sample Size
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9194
|
|
|
Intermediate Calculations
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Sample Proportion
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0.214596476
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Standard Error
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0.004171649
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Z Test Statistic
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3.498970451
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Two-Tail Test
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|
Lower Critical Value
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-1.959963985
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Upper Critical value
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1.959963985
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p-Value
|
0.000467058
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Reject the null hypothesis
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|
3 Test their claim that the true proportion of green M&Ms® candies is 0.16 at the 0.05 significance level
Details
Z Test of Hypothesis for the Proportion
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|
|
Data
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Null Hypothesis p=
|
0.16
|
Level of Significance
|
0.05
|
Number of Successes
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1684
|
Sample Size
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9194
|
|
|
Intermediate Calculations
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Sample Proportion
|
0.183162932
|
Standard Error
|
0.003823379
|
Z Test Statistic
|
6.058235535
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|
|
Two-Tail Test
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|
Lower Critical Value
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-1.959963985
|
Upper Critical value
|
1.959963985
|
p-Value
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1.37623E-09
|
Reject the null hypothesis
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|
4 Test their claim that the true proportion of yellow M&Ms® candies is 0.14 at the 0.05 significance level.
Details
Z Test of Hypothesis for the Proportion
|
|
|
Data
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Null Hypothesis p=
|
0.14
|
Level of Significance
|
0.05
|
Number of Successes
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1201
|
Sample Size
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9194
|
|
|
Intermediate Calculations
|
Sample Proportion
|
0.130628671
|
Standard Error
|
0.00361877
|
Z Test Statistic
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-2.589644785
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|
|
Two-Tail Test
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|
Lower Critical Value
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-1.959963985
|
Upper Critical value
|
1.959963985
|
p-Value
|
0.009607501
|
Reject the null hypothesis
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|
5 Test their claim that the true proportion of red M&Ms® candies is 0.13 at the 0.05 significance level.
Details
Z Test of Hypothesis for the Proportion
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|
|
Data
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Null Hypothesis p=
|
0.13
|
Level of Significance
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0.05
|
Number of Successes
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1196
|
Sample Size
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9194
|
|
|
Intermediate Calculations
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Sample Proportion
|
0.130084838
|
Standard Error
|
0.00350735
|
Z Test Statistic
|
0.024188618
|
|
|
Two-Tail Test
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|
Lower Critical Value
|
-1.959963985
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Upper Critical value
|
1.959963985
|
p-Value
|
0.980702157
|
Do not reject the null hypothesis
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|
6 Test their claim that the true proportion of brown M&Ms® candies is 0.13 at the 0.05 significance level.
Details
Z Test of Hypothesis for the Proportion
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|
|
Data
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Null Hypothesis p=
|
0.13
|
Level of Significance
|
0.05
|
Number of Successes
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1173
|
Sample Size
|
9194
|
|
|
Intermediate Calculations
|
Sample Proportion
|
0.127583206
|
Standard Error
|
0.00350735
|
Z Test Statistic
|
-0.689065494
|
|
|
Two-Tail Test
|
|
Lower Critical Value
|
-1.959963985
|
Upper Critical value
|
1.959963985
|
p-Value
|
0.490782053
|
Do not reject the null hypothesis
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|
7 On average, they claim that a 1.69 oz bag will contain more than 54 candies. Test this claim (µ > 54) at the 0.01 significance (σ unknown).
Details
Z Test of Hypothesis for the Mean
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|
|
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Data
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Null Hypothesis m=
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54
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Level of Significance
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0.01
|
Population Standard Deviation
|
2.364854141
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Sample Size
|
165
|
Sample Mean
|
55.72121212
|
|
|
Intermediate Calculations
|
Standard Error of the Mean
|
0.184103645
|
Z Test Statistic
|
9.34914743
|
|
|
Upper-Tail Test
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|
Upper Critical Value
|
2.326347874
|
p-Value
|
0
|
Reject the null hypothesis
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|
8 It is important that the total number of candies per bag does not vary very much. As a result of this quality control, the desired standard deviation is 1.5. Test the claim (α = 0.05) that the true standard deviation for number of candies per 1.69 oz bag is no more than 1.5 (σ < 1.5).
Details
Chi-Square Test of Variance
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|
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Data
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|
Null Hypothesis s^2=
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2.25
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Level of Significance
|
0.05
|
Sample Size
|
165
|
Sample Standard Deviation
|
2.364854141
|
|
|
Intermediate Calculations
|
Degrees of Freedom
|
164
|
Half Area
|
0.025
|
Chi-Square Statistic
|
407.6336701
|
|
|
Lower-Tail Test
|
|
Lower Critical Value
|
135.3900103
|
p-Value
|
1
|
Do not reject the null hypothesis
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|
Attachment:- Histogram.rar