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Sergio039 [100]
3 years ago
9

Here is data on the percentage of 200 hotels in each of the three large cities across the world on whether minibar charges are c

orrectly 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 _________.
Mathematics
1 answer:
zheka24 [161]3 years ago
5 0

Answer:

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

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