1. Lisa is a regional manager for a restaurant chain that has locations in the towns of Berwick, Milton, and Leesburg. She would like to investigate if a difference exists in the proportion of customers who rate their experience as satisfactory or better between the three locations. The following data represent the number of customers who indicated they were satisfied from random samples taken at each location.
Berwick Milton Leesburg
Number Satisfied 80 85 60
Sample Size 100 120 80
The expected frequency of satisfied customers from the Berwick sample is____.
a. 60
b. 75
c. 80
d. 90
2. Lisa is a regional manager for a restaurant chain that has locations in the towns of Berwick, Milton, and Leesburg. She would like to investigate if a difference exists in the proportion of customers who rate their experience as satisfactory or better between the three locations. The following data represent the number of customers who indicated they were satisfied from random samples taken at each location.
Berwick Milton Leesburg
Number Satisfied 80 85 60
Sample Size 100 120 80
The expected frequency of satisfied customers from the Milton sample is.
a. 60
b. 75
c. 80
d. 90
3. Lisa is a regional manager for a restaurant chain that has locations in the towns of Berwick, Milton, and Leesburg. She would like to investigate if a difference exists in the proportion of customers who rate their experience as satisfactory or better between the three locations. The following data represent the number of customers who indicated they were satisfied from random samples taken at each location.
Berwick Milton Leesburg
Number Satisfied 80 85 60
Sample Size 100 120 80
The expected frequency of satisfied customers from the Leesburg sample is.
a. 60
b. 75
c. 80
d. 90
4. Lisa is a regional manager for a restaurant chain that has locations in the towns of Berwick, Milton, and Leesburg. She would like to investigate if a difference exists in the proportion of customers who rate their experience as satisfactory or better between the three locations. The following data represent the number of customers who indicated they were satisfied from random samples taken at each location.
Berwick Milton Leesburg
Number Satisfied 80 85 60
Sample Size 100 120 80
The chi-square test statistic for these samples is_.
a. 1.49
b. 2.44
c. 4.15
d. 5.33
5. Lisa is a regional manager for a restaurant chain that has locations in the towns of Berwick, Milton, and Leesburg. She would like to investigate if a difference exists in the proportion of customers who rate their experience as satisfactory or better between the three locations. The following data represent the number of customers who indicated they were satisfied from random samples taken at each location.
Berwick Milton Leesburg
Number Satisfied 80 85 60
Sample Size 100 120 80
The degrees of freedom for the chi-square critical value is_.
a. 1
b. 2
c. 3
d. 4
6. Lisa is a regional manager for a restaurant chain that has locations in the towns of Berwick, Milton, and Leesburg. She would like to investigate if a difference exists in the proportion of customers who rate their experience as satisfactory or better between the three locations. The following data represent the number of customers who indicated they were satisfied from random samples taken at each location.
Berwick Milton Leesburg
Number Satisfied 80 85 60
Sample Size 100 120 80
The chi-square critical value using alpha = 0.05 is___.
a. 2.706
b. 3.841
c. 5.991
d. 7.815
7. Lisa is a regional manager for a restaurant chain that has locations in the towns of Berwick, Milton, and Leesburg. She would like to investigate if a difference exists in the proportion of customers who rate their experience as satisfactory or better between the three locations. The following data represent the number of customers who indicated they were satisfied from random samples taken at each location.
Berwick Milton Leesburg
Number Satisfied 80 85 60
Sample Size 100 120 80
Using alpha = 0.05, the conclusion for this chi-square test would be that because the test statistic is
A. More than the critical value, we can reject the null hypothesis and conclude that there is a difference in proportion of satisfied customers between these three locations.
B. Less than the critical value, we can reject the null hypothesis and conclude that there is a difference in proportion of satisfied customers between these three locations.
C. More 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.
D. 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.

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!!!