Question

In: Statistics and Probability

Sam has just moved from the relatively rainy city of Seattle, WA to the relatively dry...

Sam has just moved from the relatively rainy city of Seattle, WA to the relatively dry city of Phoenix, AZ. In Seattle, he found that most people preferred indoor activities over outdoor activities and wonders if this is different in Phoenix. If 61% of Seattle residents preferred indoor activities, does Phoenix significantly differ? Sam wants to test this using an alpha of .05 and the following results from his survey of Phoenix residents.

Residents Preferring Indoor Activities	Residents Preferring Outdoor Activities
49	126

1. Compute the expected frequency for Phoenix residents preferring indoor activities. Must explain/show how you arrived at your expected frequency value for credit.

2. Compute the expected frequency for Phoenix residents preferring outdoor activities. Must explain/show how you arrived at your expected frequency value for credit.

3. State your critical value for this test. Also, briefly state how you arrived at that critical value. [3 line maximum]

Expert Solution

Given:

The results from his survey of Phoenix residents.

Residents Preferring Indoor Activities	Residents Preferring Outdoor Activities
49	126

The number of Seattle residents preferred indoor activities = 61%

So the number of Seattle residents prepared outdoor activities = (100-61) = 39

Significance level, = 0.05

1) The expected frequency for Phoenix residents preferring indoor activities :

Expected frequency = 0.61 × (49 + 126) = 0.61 × 175 = 106.75

Therefore the expected frequency for Phoenix residents preferring indoor activities is 106.75

2) The expected frequency for Phoenix residents preferring outdoor activities :

Expected frequency = 0.39 × (49 + 126) = 0.39 × 175 = 68.25

Therefore the expected frequency for Phoenix residents preferring outdoor activities is 68.25

3) The critical value for this test :

We find the critical value from Chi square table.

Here, degree of freedom, df = k-1 = 2-1 = 1

At 0.05 significance level, the critical value of Chi square with degree of freedom , df = 1 is 3.8415

So 2critical = 3.8415

orchestra answered 2 years ago

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