Answer:
1) Option B is correct.
Expected frequency of satisfied customers from the Berwick sample = 75
2) Option D is correct.
Expected frequency of satisfied customers from the Milton sample = 90
3) Option A is correct.
Expected frequency of satisfied customers from the Leesburg sample = 60
4) Option B is correct.
The chi-square test statistic for these samples = 2.44
5) Option B is correct.
The degrees of freedom for the chi-square critical value = 2
6) Option C is correct.
The chi-square critical value using alpha = 0.05 is 5.991
7) Option D is correct.
The conclusion for this chi-square test would be that because the test statistic is less than the critical value, we fail to reject the null hypothesis and conclude that there is no difference in proportion of satisfied customers between these three locations.
Step-by-step explanation:
Berwick Milton Leesburg
Number Satisfied 80 85 60
Unsatisfied 20 35 20
Sample Size 100 120 80
Since this is a chi test that aims to check if there are differences in the proportion of expected number of customers for each location, we state the null and alternative hypothesis first.
The null hypothesis usually counters the claim we hope to test and would be that there is no difference between the proportion of expected frequency of satisfied customers at the three locations.
The alternative hypothesis confirms the claim we want to test and is that there is a significant difference between the proportion of expected frequency of satisfied customers at the three locations.
So, the total proportion of satisfied customers is used to calculate the expected number of satisfied customers for each of the three locations.
80+85+60= 225
Total number of customers = 100 + 120 + 80 = 300
Proportion of satisfied customers = (225/300) = 0.75
1) Expected frequency of satisfied customers from the Berwick sample = np = 100 × 0.75 = 75
2) Expected frequency of satisfied customers from the Milton sample = np = 120 × 0.75 = 90
3) Expected frequency of satisfied customers from the Leesburg sample = np = 80 × 0.75 = 60
4) Berwick Milton Leesburg
Number Satisfied 80 85 60
Unsatisfied 20 35 20
Sample Size 100 120 80
Proportion for unsatisfied ccustomers = 0.25
So, expected number of unsatisfied customers for the three branches are 25, 30 and 20 respectively.
Chi square test statistic is a sum of the square of deviations from the each expected value divided by the expected value.
χ² = [(X₁ - ε₁)²/ε₁] + [(X₂ - ε₂)²/ε₂] + [(X₃ - ε₃)²/ε₃] + [(X₄ - ε₄)²/ε₄] + [(X₅ - ε₅)²/ε₅] + [(X₆ - ε₆)²/ε₆]
X₁ = 80, ε₁ = 75
X₂ = 85, ε₂ = 90
X₃ = 60, ε₃ = 60
X₄ = 20, ε₄ = 25
X₅ = 35, ε₅ = 30
X₆ = 20, ε₆ = 20
χ² = [(80 - 75)²/75] + [(85 - 90)²/90] + [(60 - 60)²/60] + [(20 - 25)²/25] + [(35 - 30)²/30] + [(20 - 20)²/20]
χ² = 0.3333 + 0.2778 + 0 + 1 + 0.8333 + 0 = 2.4444 = 2.44
5) The degree of freedom for a chi-square test is
(number of rows - 1) × (number of columns - 1)
= (2 - 1) × (3 - 1) = 1 × 2 = 2
6) Using the Chi-square critical value calculator for a degree of freedom of 2 and a significance level of 0.05, the chi-square critical value is 5.991.
7) Interpretation of results.
If the Chi-square test statistic is less than the critical value, we fail to reject the null hypothesis.
If the Chi-square test statistic is unusually large and is greater than the critical value, we reject the null hypothesis.
For this question,
Chi-square test statistic = 2.44
Critical value = 5.991
2.44 < 5.991
test statistic < critical value
The test statistic is Less than the critical value, we fail to reject the null hypothesis and conclude that there is no difference in proportion of satisfied customers between these three locations.
Hope this Helps!!!