In: Statistics and Probability
Research suggests that children who grow up with pets are likely
to be healthier as adults because they are build up resistance to
allergens. Suppose we want to test this theory by asking a sample
of n = 40 people whether (a) they had pets during
childhood and (b) they have allergies today.
Below is the data from the study:
Pets | No Pets | ||
Allergies | ƒo = 9 | ƒo = 10 | 19 |
No Allergies | ƒo = 16 | ƒo = 5 | 21 |
25 | 15 | n = 40 |
1. Compute df and determine the critical value for χ2 for a hypothesis test with α = 0.01.
df =
Critical value of χ2 =
2. Determine the expected frequencies (ƒe) for each of
the cells in the table: (Use 3 decimals)
<
Pets | No Pets | ||
Allergies | ƒe = | ƒe = | 19 |
No Allergies | ƒe = | ƒe = | 21 |
25 | 15 | n = 40 |
3. Calculate the chi-square statistic
(χ2): (Use 3 decimals)
χ2 =
4. What decision should be made?
Retain the null.
Reject the null.
(1) The degrees of freedom, df = (r – 1) * (c -1) = (2 - 1) * (2 - 1) = 1
The Critical Value: The critical value at = 0.01 ,df = 1
critical = 6.635
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(2) The Expected value data are in the table below. Each Cell = Row total * Column Total/N. N = 40
Expected | |||
Pets | No Pets | Total | |
Allergies | 19.250 | 13.750 | 33 |
No Allergies | 15.750 | 11.250 | 27 |
Total | 35 | 25 | 60 |
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(3) The Test Statistic:
Number | Observed | Expected | (O-E) | (O-E)2 | (O-E)2/E |
1 | 15 | 19.25 | -4.25 | 18.063 | 0.9383 |
2 | 20 | 15.75 | 4.25 | 18.063 | 1.1468 |
3 | 18 | 13.75 | 4.25 | 18.063 | 1.3136 |
4 | 7 | 11.25 | -4.25 | 18.063 | 1.6056 |
Total | 5.0043 |
test = 5.004
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(4) Since test is >< c, We Retain The Null
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