Question 15. [Maximum mark: 7]
The number of coffees sold per hour at an independent coffee shop is modelled by a Poisson distribution with a mean of 22 coffees per hour.
Sheila, the shop’s owner wants to increase the number of coffees sold in the shop. She decides to offer a discount to customers who buy more than one coffee.
To test how successful this strategy is, Sheila records the number of coffees sold over a single 5-hour period. Sheila decides to use a 5 % level of significance in her test.
a. State the null and alternative hypotheses for the test. [1]
b. Find the probability that Sheila will make a type I error in her test conclusion. [4] Sheila finds 126 coffees were sold during the 5-hour period.
c. State Sheila’s conclusion to the test. Justify your answer. [2]
▶️Answer/Explanation
(a) \(H_{0} : m = 110\) , \(H_{1} : m > 110\)
(b) \(P(X ≥ 128)=0.05024 \)
\(P(X ≥ 129)= 0.04153 \) (probability of making a type I error is) 0.0415
(c) \(X ~ Po (110)\)
\(P(X≥ 126) = 0.072 > 0.05\) OR recognizing 126 < 129 or ≤ 128 so there is insufficient evidence to reject \(H_{0}\) (ie there is insufficient evidence to suggest that the number of coffees being sold has increased)
Question
The sex of cuttlefish is difficult to determine visually, so it is often found by weighing the cuttlefish.
The weights of adult male cuttlefish are known to be normally distributed with mean
10kg and standard deviation 0.5kg.
The weights of adult female cuttlefish are known to be normally distributed with mean 12 kg
and standard deviation 1 kg.
A zoologist uses the null hypothesis that in the absence of information a cuttlefish is male.
If the weight is found to be above 11.5kg the cuttlefish is classified as female.
(a) Find the probability of making a Type I error when weighing a male cuttlefish.
(b) Find the probability of making a Type II error when weighing a female cuttlefish.
90% of adult cuttlefish are male.
(c) Find the probability of making an error using the zoologist’s method.
▶️Answer/Explanation
Ans:
(a) P( ) P( Type I error stating = female when male)
\(=P(W_{male}>11.5)\)
= 0.00135 (0.00134996…)
(b) P( ) P( Type II error stating = male when female)
P 11.5) ( \(= < W_{Female} < 11.5)\)
= 0.309 (0.308537…)
(c) attempt to use the total probability
\(P(error) = 0.9 \times 0.00134996… + 0.1 \times 0.308537…\)
= 0.0321 (0.0320687…)
Question
The number of cars arriving at a junction in a particular town in any given minute between
9:00 am and 10:00 am is historically known to follow a Poisson distribution with a mean of
5.4 cars per minute.
A new road is built near the town. It is claimed that the new road has decreased the number
of cars arriving at the junction.
To test the claim, the number of cars, X, arriving at the junction between 9:00 am and
10:00 am on a particular day will be recorded. The test will have the following hypotheses:
\(H_0\) : the mean number of cars arriving at the junction has not changed,
\(H_1\) : the mean number of cars arriving at the junction has decreased.
The alternative hypothesis will be accepted if X ≤ 300.
(a) Assuming the null hypothesis to be true, state the distribution of X.
(b) Find the probability of a Type I error.
(c) Find the probability of a Type II error, if the number of cars now follows a Poisson
distribution with a mean of 4.5 cars per minute.
▶️Answer/Explanation
Ans:
(a) \(X \sim Po(324)\)
(b) \(P(X \leq 300)\)
\(=0.0946831… \approx 0.0947\)
(c) (mean number of cars =) \(4.5 \times 60 = 270\)
\(P(X > 300| \lambda = 270)\)
\(P(X \geq 301)\) OR \(1-P(X \leq 300)\)
\(=0.0334207… \approx 0.0334\)
Question
Bottles of iced tea are supposed to contain 500 ml. A random sample of 8 bottles was selected and the volumes measured (in ml) were as follows:
497.2, 502.0, 501.0, 498.6, 496.3, 499.1, 500.1, 497.7 .
a. (i) Calculate unbiased estimates of the mean and variance.
(ii) Test at the \(5\%\) significance level the null hypothesis \({{\rm{H}}_0}:\mu = 500\) against the alternative hypothesis \({{\rm{H}}_1}:\mu < 500\) .[5]
b.A random sample of size four is taken from the distribution N(60, 36) .
Calculate the probability that the sum of the sample values is less than 250.[6]
▶️Answer/Explanation
Markscheme
(i) 497.2, 502.0, 501.0, 498.6, 496.3, 499.1, 500.1, 497.7
using the GDC
\(\overline x = 499.0\) , \({\sigma ^2} = 3.8(0)\) A1A1
Note: Accept \(499\).
(ii) EITHER
\(p\)-value = 0.0950 A1
since \(0.0950 > 0.05\) accept \({H_0}\) R1A1
OR
\({t_{calc}} = – 1.45\) , \({t_{critical}} = – 1.895\) for \(v = 7\) at 5 % level A1
since \({t_{calc}} > {t_{critical}}\) accept \({H_0}\) R1A1
[5 marks]
each \(X \sim {\rm{N}}(60,36)\) so \(\sum\limits_{n = 1}^4 {{X_n} \sim {\rm{N}}(4(60),4(36)) = {\rm{N}}(240,144)} \) M1A1A1
\({\rm{Pr}}({\rm{Sum}} < 250) = {\rm{Pr}}\left( {z < \frac{{250 – 240}}{{12}} = \frac{5}{6}} \right)\) (M1)(A1)
\( = 0.798\) (by GDC) A1
Notes: Accept \(0.797\) (tables).
Answer only is awarded M0A0A0(M1)(A1)A1.
[6 marks]
Question
Bill buys two biased coins from a toy shop.
The shopkeeper claims that when one of the coins is tossed, the probability of obtaining a head is \(0.6\). Bill wishes to test this claim by tossing the coin \(250\) times and counting the number of heads obtained.
a. (i) State suitable hypotheses for this test.
(ii) He obtains \(140\) heads. Find the \(p\)-value of this result and determine whether or not it supports the shopkeeper’s claim at the \(5\%\) level of significance.[6]
b.Bill tosses the other coin a large number of times and counts the number of heads obtained. He correctly calculates a \(95\%\) confidence interval for the probability that when this coin is tossed, a head is obtained. This is calculated as [\(0.35199\), \(0.44801\)] where the end-points are correct to five significant figures.
Determine
(i) the number of times the coin was tossed;
(ii) the number of heads obtained.[7]
▶️Answer/Explanation
Markscheme
(i) \({{\rm{H}}_0}:p = 0.6\) ; \({{\rm{H}}_1}:p \ne 0.6\) A1A1
(ii) EITHER
using a normal approximation, \(p\)-value \( = 0.197\) A2
Note: Award A1 for \(0.0984\).
the shopkeeper’s claim is supported A1
because \(0.197 > 0.05\) R1
OR
using binomial distribution, \(p\)-value \( = 0.221\) A2
Note: Award A1 for \(0.110\).
the shopkeeper’s claim is supported A1
because \(0.221 > 0.05\) R1
Note: Follow through the candidate’s \(p\)-value for A1R1.
Note: Accept \(p\)-values correct to two significant figures.
[6 marks]
(i) \(\hat p = \frac{{0.35199 + 0.44801}}{2} = 0.4\) A1
width of CI \( = 3.92\sqrt {\frac{{0.4 \times 0.6}}{n}} \) M1
\(3.92\sqrt {\frac{{0.4 \times 0.6}}{n}} = 0.44801 – 0.35199 = 0.096(02)\) A1
solving,
\(n = {\left( {\frac{{3.92}}{{0.096(02)}}} \right)^2} \times 0.24\) (M1)
\( = 400\) A1
(ii) \(x = n\widehat p = 400 \times 0.4 = 160\) M1A1
[7 marks]