Find the mean and variance for the sample data given in the table.

Hence state why Aimmika believes her data follows a Poisson distribution.

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.

Find the value of $a$, of $b$, and of $c$. Give your answers to $3$ decimal places.

Write down the number of degrees of freedom for her test.

Perform the ${\chi }^2$ goodness of fit test and state, with reason, a conclusion.

By finding a critical value, perform this test at a $5 \%$ significance level.

Hence state the probability of a Type I error for this test.

By considering the claims of both Aimmika and Nichakarn, explain whether the advertising was beneficial to the store.

mean $=4.23 \left(4.23333\dots \right)$           A1

variance $=4.27 \left(4.26777\dots \right)$           A1

[2 marks]

mean is close to the variance          A1

[1 mark]

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 A1 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]

attempt to find Poisson probabilities and multiply by $90$           (M1)

$a=7.018$          A1

$b=17.498$          A1

EITHER

$90\times \text{P}\left(X\ge 8\right)=90\times \left(1-\text{P}\left(X\le 7\right)\right)$           (M1)

$c=5.755$          A1

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]

$7$        A1

[1 mark]

${\text{H}}_0 \text{:}$ The number of bags of rice sold each day follows a Poisson distribution with mean $4.2$.            A1

${\text{H}}_1 \text{:}$ 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 $\le 1$ and $\ge 8$            (M1)

$p$-value $=0.728 \left(0.728100\dots \right)$            A2

$0.728 \left(0.728100\dots \right)>0.05$            R1

the result is not significant so there is no reason to reject ${\text{H}}_0$ (the number of bags sold each day follows a Poisson distribution)            A1

Note: Do not award R0A1. The conclusion MUST follow through from their hypotheses. If no hypotheses are stated, the final A1 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]

METHOD 1

evidence of multiplying $4.2\times 60$ (seen anywhere)            M1

${\text{H}}_0 \text{:} \mu =252$

${\text{H}}_1 \text{:} \mu >252$            A1

Note: Accept ${\text{H}}_0 \text{:} \mu =4.2$ and ${\text{H}}_1 \text{:} \mu >4.2$ for the A1.

evidence of finding probabilities around critical region            (M1)

Note: Award (M1) for any of these values seen:

   $\text{P}\left(X\ge 277\right)=0.0630518\dots$  OR  $\text{P}\left(X\le 276\right)=0.936948\dots$

   $\text{P}\left(X\ge 278\right)=0.0558415\dots$  OR  $\text{P}\left(X\le 277\right)=0.944158\dots$

   $\text{P}\left(X\ge 279\right)=0.0493055\dots$  OR  $\text{P}\left(X\le 278\right)=0.950694\dots$

critical value $=279$            A1

$282\ge 279$ ,            R1

the null hypothesis is rejected            A1

(the advertising increased the number of bags sold during the $60$ days)

Note: Do not award R0A1. Accept statements referring to the advertising being effective for A1 as long as the R mark is satisfied. For the R1A1, follow through within the part from their critical value.

METHOD 2

evidence of dividing $282$ by $60$ (or $4.7$ seen anywhere)            M1

${\text{H}}_0 \text{:} \mu =4.2$

${\text{H}}_1 \text{:} \mu >4.2$            A1

attempt to find critical value using central limit theorem             (M1)

(e.g. sample standard deviation $=\sqrt{\frac{4.2}{60}}, \overline{X}~N\left(4.2, \sqrt{\frac{4.2}{60}}\right)$, etc.)

Note: Award (M1) for a $p$-value of $0.0293907\dots$ seen.

critical value $=4.63518\dots$            A1

$4.7>4.63518\dots$            R1

the null hypothesis is rejected            A1

(the advertising increased the number of bags sold during the $60$ days)

Note: Do not award R0A1. Accept statements referring to the advertising being effective for A1 as long as the R mark is satisfied. For the R1A1, follow through within the part from their critical value.

[6 marks]

$\left(\text{P}\left(X\ge 279 \overline{) \mu =252}\right)=\right) 0.0493 \left(0.0493055\dots \right)$            A1

Note: If a candidate uses METHOD 2 in part (e)(i), allow an FT answer of $0.05$ for this part but only if the candidate has attempted to find a $p$-value.

[1 mark]

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)
$14850<18000$           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.

[3 marks]

Candidates generally did well in finding the mean, although some wasted time by calculating it by hand rather than by using their GDC. Many candidates were able to find a correct variance. However, there were also many who gave the standard deviation as their variance or simply made the variance the same as their mean without performing a calculation, possibly looking ahead to part (a)(ii). Many candidates successfully used the clue given by the command term “hence” and correctly answered (a)(ii).

It was clear candidates understood that independence was the key term needed in the response. However, a number of candidates struggled either by not being precise enough in their responses (e.g. simply stating “they are independent”) or by incorrectly stating that the bags of rice sold must be independent of the number of days. “Communication” is an assessment objective for the course, and candidates should aim for clarity in their responses thereby ensuring the examiner can be confident in awarding credit.

This question was generally done well by the candidates. The two most common mistakes both stemmed from candidates not paying attention to the instructions given. Candidates either did not give their answer correct to 3 decimal places or they incorrectly attempted to use the normal distribution to find the expected frequencies. Another frequent mistake involved candidates multiplying their probabilities by 100 rather than by 90.

While most candidates were able to gain the mark in part (i), many candidates arrived at a correct answer but by using the incorrect 𝜒² test for independence method of determining the degrees of freedom. In part (ii), several common mistakes led to very few candidates receiving the full seven marks for this question part. Candidates struggled to correctly write the hypotheses as they either wrote them in reverse order or they did not correctly reference the data and/or a Poisson distribution with mean 4.2. Unfortunately, some did not state the hypotheses at all. Another common mistake involved candidates incorrectly combining columns to create a column for days when at least 7 bags of rice were sold, possibly from incorrectly thinking that an observed value less than 5 is not allowed when carrying out goodness of fit test. Many candidates also missed out on possible follow through marks for their p-value by not fully writing out the observed and expected values they were inputting into their GDC. Although communication was not being assessed here, it highlights how it is easier to credit a correct method (even if leading to an incorrect answer) if there is appropriate working and/or a running commentary present.

In contrast to part 1(d) where a method was given, many candidates struggled to know how to find a critical value in this question part. Although it was possible for candidates to find a critical value by either using Poisson probabilities or probabilities from a normal approximation, few knew how to begin. As a result, very few candidates scored full marks here.

Very few candidates managed to get full marks in this question part. Again, without being guided into a particular method, candidates struggled to understand how to begin the problem. Many candidates did not realize that some calculations were necessary to give a proper conclusion. A subset of these candidates thought it was a continuation of part (e) and made some statement related to their conclusion from the hypothesis test. For the candidates that did try to make some calculations, many simply calculated the profit from selling 282 bags of rice and compared that to the cost of the advertising. These candidates did not realize they needed to compare the difference in the expected profit without advertising and the actual profit with advertising.