Question
Aimmika is the manager of a grocery store in Nong Khai. She is carrying out a statistical analysis on the number of bags of rice that are sold in the store each day. She collects the following sample data by recording how many bags of rice the store sells each day over a period of 90 days.
She believes that her data follows a Poisson distribution.
Aimmika knows from her historic sales records that the store sells an average of 4.2 bags of rice each day. The following table shows the expected frequency of bags of rice sold each day during the 90 day period, assuming a Poisson distribution with mean 4.2 .
Aimmika decides to carry out a $\chi^2$ goodness of fit test at the $5 \%$ significance level to see whether the data follows a Poisson distribution with mean
4. 2 .
Aimmika claims that advertising in a local newspaper for 300 Thai Baht (THB) per day will increase the number of bags of rice sold. However, Nichakarn, the owner of the store, claims that the advertising will not increase the store’s overall profit.
Nichakarn agrees to advertise in the newspaper for the next 60 days. During that time, Aimmika records that the store sells 282 bags of rice with a profit of $495 \mathrm{THB}$ on each bag sold.
Aimmika wants to carry out an appropriate hypothesis test to determine whether the number of bags of rice sold during the 60 days increased when compared with the historic sales records.
a.i. Find the mean and variance for the sample data given in the table.
a.ii.Hence state why Aimmika believes her data follows a Poisson distribution.
b. State one assumption that Aimmika needs to make about the sales of bags of rice to support her belief that it follows a Poisson distribution.
c. Find the value of $a$, of $b$, and of $c$. Give your answers to 3 decimal places.
d.i. Write down the number of degrees of freedom for her test.
d.iiPerform the $\chi^2$ goodness of fit test and state, with reason, a conclusion.
e.i. By finding a critical value, perform this test at a $5 \%$ significance level.
e.ii.Hence state the probability of a Type I error for this test.
f. By considering the claims of both Aimmika and Nichakarn, explain whether the advertising was beneficial to the store.
▶️Answer/Explanation
a.i. mean $=4.23(4.23333 \ldots)$
A1
variance $=4.27(4.26777 \ldots)$
A1
[2 marks]
a.ii.mean is close to the variance
A1
[1 mark]
b. One of the following:
the number of bags sold each day is independent of any other day the sale of one bag is independent of any other bag sold
the sales of bags of rice (each day) occur at a constant mean rate
A1
Note: Award $\mathbf{A 1}$ for a correct answer in context. Any statement referring to independence must refer to either the independence of each bag sold or the independence of the number of bags sold each day. If the third option is seen, the statement must refer to a “constant mean” or “constant average”. Do not accept “the number of bags sold each day is constant”.
[1 mark]
c. attempt to find Poisson probabilities and multiply by 90
(M1)
$$
\begin{aligned}
& a=7.018 \\
& b=17.498
\end{aligned}
$$
A1
A1
EITHER
$$
\begin{aligned}
& 90 \times \mathrm{P}(X \geq 8)=90 \times(1-\mathrm{P}(X \leq 7)) \quad \text { (M1) } \\
& c=5.755 \quad \text { A1 }
\end{aligned}
$$
OR
$90-7.018-11.903-16.665-17.498-14.698-10.289-6.173$
(M1)
$$
c=5.756
$$
A1
Note: Do not penalize the omission of clear $a, b$ and $c$ labelling as this will be penalized later if correct values are interchanged.
[5 marks]
d.i. 7
A1
[1 mark]
d.ii. $\mathrm{H}_0$ : The number of bags of rice sold each day follows a Poisson distribution with mean 4.2.
A1
$\mathrm{H}_1$ : The number of bags of rice sold each day does not follow a Poisson distribution with mean 4. 2.
A1
Note: Award A1A1 for both hypotheses correctly stated and in correct order. Award A1A0 if reference to the data and/or “mean 4. 2” is not included in the hypotheses, but otherwise correct.
evidence of attempting to group data to obtain the observed frequencies for $\leq 1$ and $\geq 8$
(M1)
$p$-value $=0.728(0.728100 \ldots)$
A2
$0.728(0.728100 \ldots)>0.05$
R1
the result is not significant so there is no reason to reject $\mathrm{H}_0$ (the number of bags sold each day follows a Poisson distribution)
A1
Note: Do not award ROA1. The conclusion MUST follow through from their hypotheses. If no hypotheses are stated, the final $\mathbf{A 1}$ can still be awarded for a correct conclusion as long as it is in context (e.g. therefore the data follows a Poisson distribution).
[7 marks]
e.i. METHOD 1
evidence of multiplying $4.2 \times 60$ (seen anywhere)
M1
$$
\begin{aligned}
& \mathrm{H}_0: \mu=252 \\
& \mathrm{H}_1: \mu>252
\end{aligned}
$$
A1
Note: Accept $\mathrm{H}_0: \mu=4.2$ and $\mathrm{H}_1: \mu>4.2$ for the $\boldsymbol{A 1}$.
evidence of finding probabilities around critical region
(M1)
Note: Award (M1) for any of these values seen:
$$
\begin{aligned}
& \mathrm{P}(X \geq 277)=0.0630518 \ldots \text { OR } \mathrm{P}(X \leq 276)=0.936948 \ldots \\
& \mathrm{P}(X \geq 278)=0.0558415 \ldots \text { OR } \mathrm{P}(X \leq 277)=0.944158 \ldots \\
& \mathrm{P}(X \geq 279)=0.0493055 \ldots \text { OR } \mathrm{P}(X \leq 278)=0.950694 \ldots
\end{aligned}
$$
critical value $=279$
A1
$$
282 \geq 279 \text {, }
$$
R1
the null hypothesis is rejected
A1
(the advertising increased the number of bags sold during the 60 days)
Note: Do not award ROA1. Accept statements referring to the advertising being effective for $\boldsymbol{A 1}$ as long as the $\boldsymbol{R}$ mark is satisfied. For the $\mathrm{R1A1}$, follow through within the part from their critical value.
METHOD 2
evidence of dividing 282 by 60 (or 4.7 seen anywhere)
M1
$\mathrm{H}_0: \mu=4.2$
$\mathrm{H}_1: \mu>4.2$
A1
attempt to find critical value using central limit theorem
(M1)
(e.g. sample standard deviation $=\sqrt{\frac{4.2}{60}}, X \sim N\left(4.2, \sqrt{\frac{4.2}{60}}\right)$, etc.)
Note: Award (M1) for a $p$-value of $0.0293907 \ldots$ seen.
critical value $=4.63518 \ldots$
A1
$4.7>4.63518 \ldots$
R1
the null hypothesis is rejected
A1
(the advertising increased the number of bags sold during the 60 days)
Note: Do not award ROA1. Accept statements referring to the advertising being effective for $\boldsymbol{A 1}$ as long as the $\mathbf{R}$ mark is satisfied. For the $\mathbf{R 1 A 1}$, follow through within the part from their critical value.
[6 marks]
$$
\text { .ii. }(\mathrm{P}(X \geq 279 \mu=252)=) \quad 0.0493 \quad(0.0493055 \ldots)
$$
A1
Note: If a candidate uses METHOD 2 in part (e)(i), allow an $\boldsymbol{F T}$ answer of 0.05 for this part but only if the candidate has attempted to find a $p$-value.
[1 mark]
f. attempt to compare profit difference with cost of advertising
(M1)
Note: Award (M1) for evidence of candidate mathematically comparing a profit difference with the cost of the advertising.
EITHER
(comparing profit from 30 extra bags of rice with cost of advertising)
$$
14850<18000 \quad \text { A1 }
$$
OR
(comparing total profit with and without advertising)
$$
121590<124740
$$
A1
OR
(comparing increase of average daily profit with daily advertising cost)
$247.50<300$
A1
THEN
EITHER
Even though the number of bags of rice increased, the advertising is not worth it as the overall profit did not increase.
R1
OR
The advertising is worth it even though the cost is less than the increased profit, since the number of customers increased (possibly buying other products and/or returning in the future after advertising stops)
R1
Note: Follow through within the part for correct reasoning consistent with their comparison.