Question-67

Chi-square Test

The demand for a particular spare part in a factory was found to very from day-to-day. In a sample study the following information was obtained:

Days Mon. Tues. Wed. Thurs. Fri. Sat.
No. of parts Demanded 1124 1125 1110 1120 1126 1115

Test the hypothesis that the number of parts demanded does not depend on the day of the week. (Given : the values of chi-square significance at 5, 6, 7, d.f. are respectively 11.07, 12.59, 14.07 at 5% level of significance).

Use the formula of “Chi-square test of goodness of fit” as

\(\chi^2 = \sum_{i=1}^{n}\left[\dfrac{(f_i - e_i)^2}{e_i}\right] \sim \chi^2_{n-1}\),

where, \(f_i (i = 1, 2, \ldots, n)\) is a set of observed frequencies and \(e_i (i = 1, 2, \ldots, n)\) is the corresponding set of expected frequencies such that \(\sum_{i=1}^{n}f_i = \sum_{i=1}^{n}e_i\).

Also, expected frequencies, \(e_i\), can be calculated as

\(e_i = np_i,\) where

\(n=\) total number of observations

\(p_i =\) probability of category \(i\)

\(H_0:\) The number of parts demanded does not depend on the day of the week.

Under the null hypothesis, the expected frequencies of the spare parts demanded on each of the six days wpuld be:

\(\dfrac{1}{6}(1124+1125+1110+1120+1126+1115) = \dfrac{6720}{6} = 1120\)

Days Observed freq. \((f_i)\) Expected freq. \((e_i)\) \((fi - e_i)^2\) \(\dfrac{(fi - e_i)^2}{e_i}\)
Mon. 1124 1120 16 0.014
Tues. 1125 1120 25 0.022
Wed. 1110 1120 100 0.089
Thurs. 1120 1120 0 0
Fri. 1126 1120 36 0.032
Sat. 1115 1120 25 0.022
Total 6720 6720 0.179

Thus, \(\chi^2 = \sum_{i=1}^{n}\left[\dfrac{(f_i - e_i)^2}{e_i}\right] = 0.179\), which is less than the tablulated value of \(\chi^2_{n-1} = \chi^2_5 = 11.07\).

Thus, we will accept the null hypothesis (or reject the alternative hypothesis ) and conclude that the number of spare parts demanded does not depend on the day of the week.

Hence, option (a) and (d) are correct.