Step 3

Completing the table we have $E= \var{t}/5=\var{t/5}$ for all brands.

We do the last column calculations for brand A.

$\displaystyle \frac{(O -E) ^ 2}{ E} = \frac{(\var{a} -\var{e1}) ^ 2} {\var{e1}} = \var{x[0]}$

to 2 decimal places.

The test statistic is then:

$$\displaystyle \chi ^ 2 = \sum \frac{(O -E)^2}{E} = \var{x[0]} + \var{x[1]} + \var{x[2]} + \var{x[3]} + \var{x[4]} = \var{chi}$$

Step 4:

The degrees of freedom is given by: $\nu$= no. of categories $- 1 = 5-1=4$

The following are the critical values for $\nu=4$.

Looking at this test statistic we see that the p-range is greater than $10$%.

The conclusion we come to is that there is no evidence to suggest that customers have a preference between the brands. Hence we retain the null hypothesis.

Step 3.

a) $\lambda$  is the mean of the data i.e.

\[\lambda = \frac{0 \times \var{a}+1\times \var{b}+2\times\var{c}+3\times\var{d}+4\times\var{f}+5\times \var{t}}{\var{thismany}} = \var{v}\] to 2 decimal places.

b) This is the value of $\lambda$ we use to calculate the probabilities, to 3 decimal places,  assuming that the data is from a Poisson distribution with this parameter.

For example, $ \displaystyle P(X=2)=\frac{e^{-\lambda}\lambda^2}{2!}=\frac{e^{-\var{v}}\times \var{v}^2}{2}=\var{x[2]}$

to 3 decimal places.

The next table shows the values to 3 decimal places you should have obtained:

c) Using this probability then the expected number of occurences of 2 calls in a minute is \[P(X=2) \times \var{thismany} = \var{x[2]}\times \var{thismany}=\var{ex[2]}\]  to 2 decimal places.

In the same way we find all the expected frequencies assuming that it is a Poisson distribution with parameter $\lambda=\var{v}$.

The following table shows the actual observed frequencies and the expected frequencies, to 2 decimal places, under this assumption.

The expected frequencies in the last 2 categories need to be pooled as the  expected frequency for the last category is less than 5.

Hence we obtain the following table used to calculate the test statistic:

We now use the last table to calculate the $\chi^2$ statistic to see if there is a reasonable match between the observed and expected frequencies. 

We have:

\[\chi ^ 2 = \simplify[all,!collectNumbers]{({a} -{ex[0]}) ^ 2 / {ex[0]} + ({b} -{ex[1]}) ^ 2 / {ex[1]} + ({c} -{ex[2]}) ^ 2 / {ex[2]} + ({dp} -{ep4}) ^ 2 / {ep4} + {z} * (({fp} -{ep5}) ^ 2 / {ep5}) = {ch}}\] to 2 decimal places.

Step 4:

The degrees of freedom is given by:  $\nu=\;$no. of pooled categories $- 2 =\var{6-few+1}-2=\var{5-few}$

(We have to take away $2$ from the number of pooled categories as we have estimated the parameter $\lambda$ .)

The following are the critical values for $\nu=\var{5-few}$:

Comparing these values with the the test statistic we see that the  $p$-value is greater than 10%.

Step 5: Conclusion

Hence there is no evidence against and so we retain the null hypothesis that the number of calls per minute follows a Poisson distribution.

Step 3

The expected frequencies are given by replacing a value in the table by the expected value:

\[E = \frac{\textrm{row total} \times \textrm{column total}}{\textrm{overall total}}\]

For example, the Excellent category for Marketing & Management lies in the second row (with sum $\var{r2}$) and the first column (with sum $\var{col1}$).

So the expected frequency of Excellent Marketing & Management students is:
\[E = \simplify[]{({r2}*{col1})/({tot})} = \var{e4}\]

Hence we get the following table of expected frequencies:

In order to test to see if there is an association we compare this table with the table of observed values and calculate the test statistic by looking at

\[\chi^2 = \sum \frac{(O - E)^2}{E}\]

Calculating the values for Business Management and Mathematics, all to 3 decimal places, you should obtain:

To find the $\chi^2$ statistic you sum these fifteen values to get:

\[\begin{eqnarray} \chi^2 &=& \var{ch1} + \var{ch2} + \var{ch3} + \var{ch4} + \var{ch5} +\\ && \var{ch6} + \var{ch7} + \var{ch8} + \var{ch9} + \var{ch10} +\\ && \var{ch11} + \var{ch12} + \var{ch13} + \var{ch14} + \var{ch15} \\ &=& \var{ch}. \end{eqnarray}\]

Step 4

The degrees of freedom is given by:

\[\nu = (\textrm{no. of rows} - 1) \times (\textrm{no. of columns} - 1) = 4 \times 2 = 8\]

The following are the critical values for $\nu = 8$:

Comparing these values with the the test statistic we see that the $p$-value {pResult}.

Step 5: Conclusion

As the $p$-value {pResult}, there is {evi} evidence against $H_0$.

Hence we {retain}.

{summary}