Sample Multinomial Experiments problems
Question 1: The manager of a Farmer Jack Super Market would like to know if there is a preference for the day of the week on which customers do their shopping. A sample of 420 families revealed the following. At the 0.05 significance level, is there a difference in the proportion of customers that prefer each day of the week? Chi Square test. Goodness of Fit Equal expected frequencies.
Day of the week |
Number of persons |
Monday |
20 |
Tuesday |
30 |
Wednesday |
20 |
Thursday |
60 |
Friday |
80 |
Saturday |
130 |
Sunday |
80 |
Solution: The following null hypothesis need to be tested:
\[H_0:\,p_{1} = {1/7},\,\,\, p_{2} = {1/7},\,\,\, p_{3} = {1/7},\,\,\, p_{4} = {1/7},\,\,\, p_{5} = {1/7},\,\,\, p_{6} = {1/7},\,\,\, p_{7} = {1/7}\]
The first task is building the table with expected values. Based on the data provided, we find:
Category |
Observed |
Expected |
(fo – fe)²/fe |
Monday |
20 |
420*1/7=60 |
26.6667 |
Tuesday |
30 |
420*1/7=60 |
15 |
Wednesday |
20 |
420*1/7=60 |
26.6667 |
Thursday |
60 |
420*1/7=60 |
0 |
Friday |
80 |
420*1/7=60 |
6.6667 |
Saturday |
130 |
420*1/7=60 |
81.6667 |
Sunday |
80 |
420*1/7=60 |
6.6667 |
Sum = |
163.3333 |
This means the Chi-Square statistics is computed as
\[{{\chi }^{2}}=\sum\limits_{i=1}^{n}{\frac{{{\left( {{O}_{i}}-{{E}_{i}} \right)}^{2}}}{{{E}_{i}}}}={26.6667} + {15} + {26.6667} + {0} + {6.6667} + {81.6667} + {6.6667}=163.3333\]
The critical value for \(\alpha =0.05\) and \(df = 6\) is given by
\[\chi _{C}^{2}= {12.5916}\]
and the corresponding p-value is
\[p=\Pr \left( {{\chi }^{2}}> {163.3333} \right) = {0.000}\]
Since the p-value is less than the significance level \(\alpha = {0.05}\), then we reject \({{H}_{0}}\). This means that we have enough evidence to reject the null hypothesis of equal proportions, at the 0.05 significance level.
Question 2: Research has demonstrated that people tend to be attracted to others who are similar to themselves. One study demonstrated that individuals are disproportionately more likely to marry those with surnames that begin with the same last letter as their own (Jones, Pelham, Carvallo, & Mirenberg, 2004). The researchers began by looking at marriage records and recording the surname for each groom and the maiden name of each bride. From these records it is possible to calculate the probability of randomly matching a bride and a groom whose last names begin with the same letter. Suppose that this probability is only 6.5%. Next, a sample of n 200.married couples is selected and the number who shared the same last initial at the time they were married is counted. The resulting observed frequencies are as follows:
Do these date indicate that the number of couples with the same last initial is significantly different that would be expected if couples were matched randomly? Test with a = .05.
Solution: The following null hypothesis need to be tested:
\[H_0:\,p_{1} = {0.065},\,\,\, p_{2} = {0.935}\]
The first task is building the table with expected values. Based on the data provided, we find:
Category |
Observed |
Expected |
(fo – fe)²/fe |
Same Initial |
19 |
200*0.065=13 |
2.7692 |
Different Initials |
181 |
200*0.935=187 |
0.1925 |
Sum = |
2.9617 |
Using that information, we get
\[{{\chi }^{2}}=\sum\limits_{i=1}^{n}{\frac{{{\left( {{O}_{i}}-{{E}_{i}} \right)}^{2}}}{{{E}_{i}}}}={2.7692} + {0.1925}=2.9617\]
The critical value for \(\alpha =0.05\) and \(df = 1\) is given by
\[\chi _{C}^{2}= {3.8415}\]
and the corresponding p-value is
\[p=\Pr \left( {{\chi }^{2}}> {2.9617} \right) = {0.0853}\]
Since the p-value is greater than the significance level \(\alpha = {0.05}\), then we fail to reject \({{H}_{0}}\). This means that we don’t have enough evidence to reject the null hypothesis of the given proportions, at the 0.05 significance level.
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