Concept explainers
a)
To state hypotheses.
a)
Answer to Problem 17E
Explanation of Solution
Given:
The flavor blend is 20% of each flavor.
Calculation:
The null and alternative hypotheses:
b)
To find expected count.
b)
Answer to Problem 17E
Expected count for each flavor is 12.
Explanation of Solution
Given:
The flavor blend is 20% of each flavor.
Calculation:
For finding expected count, we need to multiply by 60 to each frequency of flavor. Therefore, expected count is,
Flavor | Frequency | Expected |
Lemon | 0.2 | 12 |
Lime | 0.2 | 12 |
Orange | 0.2 | 12 |
Strawberry | 0.2 | 12 |
Grape | 0.2 | 12 |
c)
To explain how large a test statistic would need a significant evidence against claim.
c)
Answer to Problem 17E
If chi square statistic is greater than above critical values at specified level of significance then reject H0.
Explanation of Solution
Given:
The flavor blend is 20% of each flavor.
Sample size is 60 candies.
Flavor | Frequency | Expected |
Lemon | 0.2 | 12 |
Lime | 0.2 | 12 |
Orange | 0.2 | 12 |
Strawberry | 0.2 | 12 |
Grape | 0.2 | 12 |
Formula:
The degrees of freedom = df = c-1
Calculation:
The df = 5-1 = 4
There will be different critical values for each level of significance.
When significance level = a = 0.05
Decision: If chi square statistic is greater than above critical values at specified level of significance then reject H0.
d)
To test a claim by taking own sample data for 60 candies.
d)
Explanation of Solution
Given:
The flavor blend is 20% of each flavor.
Sample size is 60 candies.
Calculation:
Following table shows the random sample and their frequency:
Flavor | Observed | Frequency |
Lemon | 20 | 0.2 |
Lime | 10 | 0.2 |
Orange | 15 | 0.2 |
Strawberry | 4 | 0.2 |
Grape | 11 | 0.2 |
Using excel,
Flavor | observed | expected | O - E | (O - E)² / E | % of chisq |
Lemon | 20 | 12.000 | 8.000 | 5.333 | 45.07 |
Lime | 10 | 12.000 | -2.000 | 0.333 | 2.82 |
Orange | 15 | 12.000 | 3.000 | 0.750 | 6.34 |
Strawberry | 4 | 12.000 | -8.000 | 5.333 | 45.07 |
Grape | 11 | 12.000 | -1.000 | 0.083 | 0.70 |
60 | 60.000 | 0.000 | 11.833 | 100.00 | |
11.83 | chi-square | ||||
4 | df | ||||
.0186 | p-value |
Therefore, the p-value is between 0.01 and 0.05
Chapter 11 Solutions
PRACTICE OF STATISTICS F/AP EXAM
Additional Math Textbook Solutions
An Introduction to Mathematical Statistics and Its Applications (6th Edition)
Introductory Statistics
Statistics: The Art and Science of Learning from Data (4th Edition)
Intro Stats
Intro Stats, Books a la Carte Edition (5th Edition)
Introductory Statistics (10th Edition)
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