In: Statistics and Probability
It has been suggusted that the highest priority of retirees is travel. Thus, a study was conducted to investigate the differences in the length of stay of a trip for pre- and post-retirees. A sample of 691 travelers were asked how long they stayed on a typical trip. The observed results of the study are found below.
Number of Nights | Pre-retirement | Post-retirement | Total |
4−7 | 235 | 175 | 410 |
8−13 | 84 | 63 | 147 |
14−21 | 31 | 59 | 90 |
22 or more | 12 | 32 | 44 |
Total | 362 | 329 | 691 |
With this information, construct a table of estimated expected
values.
Number of Nights | Pre-retirement | Post-retirement |
4−7 | ||
8−13 | ||
14−21 | ||
22 or more |
Now, with that information, determine whether the length of stay is
independent of retirement using α=0.01.
(a) χ2=
(b) Find the degrees of freedom:
(c) Find the critical value:
We have to find out Expected Frequencies:
Ei = (Corresponding row total * Corresponding column total) / sample size
For First cell:
E1 = (410 * 362) / 691 = 214.79
E2 = (410 * 329) / 691 = 195.21
Similarly solve for all other cells,
We have calculated for expected frequencies and presented in the following table:
Expected Frequencies:
Number of Nights | Pre -retirement | Post-retirement | Total |
4 - 7 | 214.79 | 195.21 | 410.00 |
8 - 13 | 77.01 | 69.99 | 147.00 |
14 - 21 | 47.15 | 42.85 | 90.00 |
22 or more | 23.05 | 20.95 | 44.00 |
Total | 362.00 | 329.00 | 691.00 |
a)
Computational Table:
Oi | Ei | (Oi-EI) | (Oi-Ei)2 | (Oi-Ei)2/Ei | |
235 | 214.79 | 20.21 | 408.4377 | 1.901566 | |
84 | 77.01 | 6.99 | 48.85828 | 0.63444 | |
31 | 47.15 | -16.15 | 260.7921 | 5.531226 | |
12 | 23.05 | -11.05 | 122.1169 | 5.297763 | |
175 | 195.21 | -20.21 | 408.4377 | 2.092301 | |
63 | 69.99 | -6.99 | 48.85828 | 0.698076 | |
59 | 42.85 | 16.15 | 260.7921 | 6.08603 | |
32 | 20.95 | 11.05 | 122.1169 | 5.82915 | |
Total | 691 | 691.00 | 28.07 |
Test statistic:
b)
Degrees of Freedom = (r-1) * (c-1)
Where, r = Number of rows = 4
c = Number of columns = 2
Degrees of Freedom = (r-1) * (c-1) = (4-1) * (2-1) = 3*1 = 3
C)
Critical value:
.........................From Chi square table
Conclusion:
Test statistic > Critical value, i.e. 28.07 > 11.34, That is Reject Ho at 1% level of significance.
Therefore, The length of stay is dependent of retirement