Here is data on the percentage of 200 hotels in each of the three large cities across the world on whether minibar charges are correctly posted at checkout are given below. Hong Kong New York Paris
Yes 86% 76% 78%
No 14% 24% 22%
At the 0.05 level of significance, you want to know if there is evidence of a difference in the proportion of hotels that correctly post minibar charges among the three cities.
Referring to the table above the test will involve degrees of freedom.
Referring to Scenario above, the expected cell frequency for the Hong Kong/Yes cell is ?
Referring to Scenario above, the critical value of the test is . Use degrees of freedom and look at the chi-square distribution table.
Referring to Scenario above, the value of the test statistic is .

Degree of freedom = 2

Expected frequency = 80%

Critical value, = 5.991

χ² statistic = 3.5

Step-by-step explanation:

Given the data :

Hong Kong New York Paris

Yes 86% 76% 78%

No 14% 24% 22%

The degree of freedom for the Chisquare statistic is given as :

(no of rows - 1) * (number of columns - 1)

Number of rows = 2

Number of columns = 3

. Degree of freedom = (2-1) * (3-1) = 1*2 = 2

Expected frequency = (Row total * column total) / grand total

The expected frequency of Hong Kong / Yes cell :

Row total = (86+76+78) = 240

Column total = (86+14) = 100

Grand total, N = (14+24+22)+240 = 300

Expected frequency = (240*100)/300 = 80%

The critical value :

At α - level = 0.05 ; df = 2

Critical value = 5.991

χ² = Σ(observed - Expected)² / Expected

The expected values :

80 80 80

20 20 20

Hence,

χ² = Σ(86-80)²/80 + (76-80)²/80 + (78-80)²/80 + (14-20)²/20 + (24-20)²/20 + (22-20)²/20

χ² statistic = 3.5

(1,-1) i hope this

step-by-step explanation: