IB DP Maths Topic 7.6 Null and alternative hypotheses, H0 and H1 HL Paper 3

 

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Question

The random variable X has a Poisson distribution with mean \(\mu \). The value of \(\mu \) is known to be either 1 or 2 so the following hypotheses are set up.

\[{{\text{H}}_0}:\mu = 1;{\text{ }}{{\text{H}}_1}:\mu = 2\]

A random sample \({x_1},{\text{ }}{x_2},{\text{ }} \ldots ,{\text{ }}{x_{10}}\) of 10 observations is taken from the distribution of X and the following critical region is defined.

\[\sum\limits_{i = 1}^{10} {{x_i} \geqslant 15} \]

Determine the probability of

(a)     a Type I error;

(b)     a Type II error.

Answer/Explanation

Markscheme

(a)     let \(T = \sum\limits_{i = 1}^{10} {{X_i}} \) so that T is Po(10) under \({{\text{H}}_0}\)     (M1)

\({\text{P(Type I error)}} = {\text{P }}T \geqslant 15|\mu = 1\)     M1A1

\( = 0.0835\)     A2     N3

Note: Candidates who write the first line and only the correct answer award (M1)M0A0A2.

 

[5 marks]

 

(b)     let \(T = \sum\limits_{i = 1}^{10} {{X_i}} \) so that T is Po(20) under \({{\text{H}}_1}\)     (M1)

\({\text{P(Type II error)}} = {\text{P }}T \leqslant 14|\mu  = 2\)     M1A1

\( = 0.105\)     A2     N3

Note: Candidates who write the first line and only the correct answer award (M1)M0A0A2.

 

Note: Award 5 marks to a candidate who confuses Type I and Type II errors and has both answers correct.

 

[5 marks]

Total [10 marks]

Examiners report

This question caused problems for many candidates and the solutions were often disappointing. Some candidates seemed to be unaware of the meaning of Type I and Type II errors. Others were unable to calculate the probabilities even when they knew what they represented. Candidates who used a normal approximation to obtain the probabilities were not given full credit – there seems little point in using an approximation when the exact value could be found.

Question

Ten friends try a diet which is claimed to reduce weight. They each weigh themselves before starting the diet, and after a month on the diet, with the following results.

 

Determine unbiased estimates of the mean and variance of the loss in weight achieved over the month by people using this diet.

[5]
a.

(i)     State suitable hypotheses for testing whether or not this diet causes a mean loss in weight.

(ii)     Determine the value of a suitable statistic for testing your hypotheses.

(iii)     Find the 1 % critical value for your statistic and state your conclusion.

[6]
b.
Answer/Explanation

Markscheme

the weight losses are

2.2\(\,\,\,\,\,\)3.5\(\,\,\,\,\,\)4.3\(\,\,\,\,\,\)–0.5\(\,\,\,\,\,\)4.2\(\,\,\,\,\,\)–0.2\(\,\,\,\,\,\)2.5\(\,\,\,\,\,\)2.7\(\,\,\,\,\,\)0.1\(\,\,\,\,\,\)–0.7     (M1)(A1)

\(\sum {x = 18.1} \), \(\sum {{x^2} = 67.55} \)

UE of mean = 1.81     A1

UE of variance \( = \frac{{67.55}}{9} – \frac{{{{18.1}^2}}}{{90}} = 3.87\)     (M1)A1

Note: Accept weight losses as positive or negative. Accept unbiased estimate of mean as positive or negative.

 

Note: Award M1A0 for 1.97 as UE of variance.

 

[5 marks]

a.

(i)     \({H_0}:{\mu _d} = 0\) versus \({H_1}:{\mu _d} > 0\)     A1

Note: Accept any symbol for \({\mu _d}\)

 

(ii)     using t test     (M1)

\(t = \frac{{1.81}}{{\sqrt {\frac{{3.87}}{{10}}} }} = 2.91\)     A1

 

(iii)     DF = 9     (A1)

Note: Award this (A1) if the p-value is given as 0.00864

 

1% critical value = 2.82     A1

accept \({H_1}\)     R1

Note: Allow FT on final R1.

 

[6 marks]

b.

Examiners report

In (a), most candidates gave a correct estimate for the mean but the variance estimate was often incorrect. Some candidates who use their GDC seem to be unable to obtain the unbiased variance estimate from the numbers on the screen. The way to proceed, of course, is to realise that the larger of the two ‘standard deviations’ on offer is the square root of the unbiased estimate so that its square gives the required result. In (b), most candidates realised that the t-distribution should be used although many were awarded an arithmetic penalty for giving either t = 2.911 or the critical value = 2.821. Some candidates who used the p-value method to reach a conclusion lost a mark by omitting to give the critical value. Many candidates found part (c) difficult and although they were able to obtain t = 2.49…, they were then unable to continue to obtain the confidence interval.

a.

In (a), most candidates gave a correct estimate for the mean but the variance estimate was often incorrect. Some candidates who use their GDC seem to be unable to obtain the unbiased variance estimate from the numbers on the screen. The way to proceed, of course, is to realise that the larger of the two ‘standard deviations’ on offer is the square root of the unbiased estimate so that its square gives the required result. In (b), most candidates realised that the t-distribution should be used although many were awarded an arithmetic penalty for giving either t = 2.911 or the critical value = 2.821. Some candidates who used the p-value method to reach a conclusion lost a mark by omitting to give the critical value. Many candidates found part (c) difficult and although they were able to obtain t = 2.49…, they were then unable to continue to obtain the confidence interval.

b.

Question

The random variable X has a Poisson distribution with unknown mean \(\mu \) . It is required to test the hypotheses

\({H_0}:\mu  = 3\) against \({H_1}:\mu  \ne 3\) .

Let S denote the sum of 10 randomly chosen values of X . The critical region is defined as \((S \leqslant 22) \cup (S \geqslant 38)\) .

Calculate the significance level of the test.

[5]
a.

Given that the value of \(\mu \) is actually 2.5, determine the probability of a Type II error.

[5]
b.
Answer/Explanation

Markscheme

under \({H_0}\) , \(S{\text{ is Po}}(30)\)     (A1)

EITHER

\({\text{P}}(S \leqslant 22) = {\text{0.080569}} \ldots \)     A1

\({\text{P}}(S \geqslant 38) = {\text{0.089012}} \ldots \)     A1

significance level = 0.080569… + 0.089012…     (M1)

= 0.170     A1

OR

\({\text{P}}(S \leqslant 22) = {\text{0.080569}} \ldots \)     A1

\({\text{P}}(S \leqslant 37) = {\text{0.910987}} \ldots \)     A1

significance level = 1 – (0.910987…) + 0.089012…     (M1)

= 0.170     A1

Note: Accept 17 % or 0.17.

 

Note: Award 2 marks out of the final 4 marks for correct use of the Central Limit Theorem, giving 0.144 without a continuity correction and 0.171 with a continuity correction. The first (A1) is independent.

 

[5 marks]

a.

S is now Po(25)     (A1)

P (Type II error) = P (accept \({H_0}|\mu = 2.5\))     (M1)

\( = {\text{P}}\left( {23 \leqslant S \leqslant 37|S{\text{ is Po}}(25)} \right)\)     (M1)

Note: Only one of the above M1 marks can be implied.

 

= 0.990789… – 0.317533…     (A1)

= 0.673     A1

Note: Award 2 marks out of the final 4 marks for correct use of the Central Limit Theorem, giving 0.647 without a continuity correction and 0.685 with a continuity correction. The first (A1) is independent.

 

[5 marks]

b.

Examiners report

Solutions to this question were often disappointing with many candidates not knowing what had to be done. Even those candidates who knew what to do sometimes made errors in evaluating the probabilities, often by misinterpreting the inequality signs. Candidates who used the Central Limit Theorem to evaluate the probabilities were given only partial credit on the grounds that the answers obtained were approximate and not exact.

a.

Solutions to this question were often disappointing with many candidates not knowing what had to be done. Even those candidates who knew what to do sometimes made errors in evaluating the probabilities, often by misinterpreting the inequality signs. Candidates who used the Central Limit Theorem to evaluate the probabilities were given only partial credit on the grounds that the answers obtained were approximate and not exact.

b.

Question

A baker produces loaves of bread that he claims weigh on average 800 g each. Many customers believe the average weight of his loaves is less than this. A food inspector visits the bakery and weighs a random sample of 10 loaves, with the following results, in grams:

783, 802, 804, 785, 810, 805, 789, 781, 800, 791.

Assume that these results are taken from a normal distribution.

Determine unbiased estimates for the mean and variance of the distribution.

[3]
a.

In spite of these results the baker insists that his claim is correct.

Stating appropriate hypotheses, test the baker’s claim at the 10 % level of significance.

[7]
b.
Answer/Explanation

Markscheme

unbiased estimate of the mean: 795 (grams)     A1

unbiased estimate of the variance: 108 \((gram{s^2})\)     (M1)A1

[3 marks]

a.

null hypothesis \({H_0}:\mu  = 800\)     A1

alternative hypothesis \({H_1}:\mu  < 800\)     A1

using 1-tailed t-test     (M1)

EITHER

p = 0.0812…     A3

OR

with 9 degrees of freedom     (A1)

\({t_{calc}} = \frac{{\sqrt {10} (795 – 800)}}{{\sqrt {108} }} = – 1.521\)     A1

\({t_{crit}} = – 1.383\)     A1 

Note: Accept 2sf intermediate results.

 

THEN

so the baker’s claim is rejected     R1 

Note: Accept “reject \({H_0}\) ” provided \({H_0}\) has been correctly stated.

 

Note: FT for the final R1.

 

[7 marks]

b.

Examiners report

A successful question for many candidates. A few candidates did not read the question and adopted a 2-tailed test.

a.

A successful question for many candidates. A few candidates did not read the question and adopted a 2-tailed test.

b.

Question

The random variable X represents the height of a wave on a particular surf beach.

It is known that X is normally distributed with unknown mean \(\mu \) (metres) and known variance \({\sigma ^2} = \frac{1}{4}{\text{ (metre}}{{\text{s}}^2}{\text{)}}\) . Sally wishes to test the claim made in a surf guide that \(\mu  = 3\) against the alternative that \(\mu  < 3\) . She measures the heights of 36 waves and calculates their sample mean \({\bar x}\) . She uses this value to test the claim at the 5 % level.

(i)     Find a simple inequality, of the form \(\bar x < A\) , where A is a number to be determined to 4 significant figures, so that Sally will reject the null hypothesis, that \(\mu  = 3\) , if and only if this inequality is satisfied.

(ii)     Define a Type I error.

(iii)     Define a Type II error.

(iv)     Write down the probability that Sally makes a Type I error.

(v)     The true value of \(\mu \) is 2.75. Calculate the probability that Sally makes a Type II error.

[11]
a.

The random variable Y represents the height of a wave on another surf beach. It is known that Y is normally distributed with unknown mean \(\mu \) (metres) and unknown variance \({\sigma ^2}{\text{ (metre}}{{\text{s}}^2}{\text{)}}\) . David wishes to test the claim made in a surf guide that \(\mu = 3\) against the alternative that \(\mu < 3\) . He is also going to perform this test at the 5 % level. He measures the heights of 36 waves and finds that the sample mean, \(\bar y = 2.860\) and the unbiased estimate of the population variance, \(s_{n – 1}^2 = 0.25\).

(i)     State the name of the test that David should perform.

(ii)     State the conclusion of David’s test, justifying your answer by giving the p-value.

(iii)     Using David’s results, calculate the 90 % confidence interval for \(\mu \) , giving your answers to 4 significant figures.

[8]
b.
Answer/Explanation

Markscheme

(i)     \({H_0}:\mu = 3,{\text{ }}{H_1}:\mu < 3\)

1 tailed z test as \({\sigma ^2}\) is known

under \({H_0}{\text{, }}X \sim {\text{N}}\left( {3,\frac{1}{4}} \right){\text{ so }}\bar X \sim {\text{N}}\left( {3,\frac{{\frac{1}{4}}}{{36}}} \right) = N\left( {3,\frac{1}{{144}}} \right)\)     (M1)

\(z = \frac{{\bar x – 3}}{{\frac{1}{{12}}}}{\text{ is N(0, 1)}}\)     (A1)

\({\text{P}}(z < – 1.64485…) = 0.05\)     (A1)

so inequality is given by \(\frac{{\bar x – 3}}{{\frac{1}{{12}}}} < – 1.64485…{\text{ giving }}\bar x < 2.8629…\)     M1

\(\bar x < 2.863{\text{ (4sf)}}\)     A1

Note: Candidates can get directly to the answer from \({\text{N}}\left( {3,\frac{1}{{144}}} \right)\) they do not have to go via z is N(0, 1) . However they must give some explanation of what they have done; they cannot just write the answer down.

 

(ii)     a Type I error is accepting \({H_1}\) when \({H_0}\) is true     A1

 

(iii)     a Type II error is accepting \({H_0}\) when \({H_1}\) is true     A1

 

(iv)     0.05     A1

Note: Accept anything that rounds to 0.050 if they do the conditional calculation.

 

(v)     \(\bar X \sim {\text{N}}\left( {2.75,\frac{1}{{144}}} \right)\)     (M1)

\({\text{P}}(\bar x > 2.8629…) = 0.0877{\text{ (3sf)}}\)     (M1)A1

Note: Accept any answer between 0.0875 and 0.0877 inclusive.

 

Note: Accept anything that rounded is between 0.087and 0.089 if there is evidence that the candidate has used tables.

 

[11 marks]

a.

(i)     t-test     A1

 

(ii)     \({{\text{H}}_0}:\mu = 3,{\text{ }}{{\text{H}}_1}:\mu < 3\)

1 tailed t test as \({\sigma ^2}\) is unknown

\(t = \frac{{\bar y – 3}}{{\frac{1}{{12}}}}\) has the t-distribution with \(v = 35\)     (M1)

the p-value is 0.0509…     A2

this is \( > 0.05\)     R1

so we accept that the mean wave height is 3     R1

Note: Allow “Accept \({{\text{H}}_0}\) ” provided \({{\text{H}}_0}\) has been stated.

 

Note: Accept FT on the p-value for the R1s.

 

(iii)     \(2.719 < \mu < 3.001{\text{ (4 sf)}}\)     A1A1

Note: \(2.860 \pm 1.6896… \times \frac{{\frac{1}{2}}}{6}\) would gain M1.

 

Note: Award A1A0 if answer are only given to 3sf.

 

[8 marks]

b.

Examiners report

(a) There were many reasonable answers. In (i) not all candidates explained their method so that they could gain good partial marks even if they had the wrong final answer. A common mistake was to give an answer above 3. It was pleasing that almost all candidates had (ii) and (iii) correct, as this had caused problems in the past. In (iv) it was amusing to see a few candidates work out 5% using conditional probability rather than just write down the answer as asked.

(b) It was pleasing that almost all candidates realised that it was a t-test rather than a z-test.

There was good understanding on how to use the calculator in parts (ii) and (iii). The correct confidence interval to the desired accuracy was not always given.

The most common mistake in question 3 was forgetting to take into account the variance of the sample mean.

a.

(a) There were many reasonable answers. In (i) not all candidates explained their method so that they could gain good partial marks even if they had the wrong final answer. A common mistake was to give an answer above 3. It was pleasing that almost all candidates had (ii) and (iii) correct, as this had caused problems in the past. In (iv) it was amusing to see a few candidates work out 5% using conditional probability rather than just write down the answer as asked.

(b) It was pleasing that almost all candidates realised that it was a t-test rather than a z-test.

There was good understanding on how to use the calculator in parts (ii) and (iii). The correct confidence interval to the desired accuracy was not always given.

The most common mistake in question 3 was forgetting to take into account the variance of the sample mean.

b.

Question

The random variable X is normally distributed with unknown mean \(\mu \) and unknown variance \({\sigma ^2}\). A random sample of 20 observations on X gave the following results.

\[\sum {x = 280,{\text{ }}\sum {{x^2} = 3977.57} } \]

Find unbiased estimates of \(\mu \) and \({\sigma ^2}\).

[3]
a.

Determine a 95 % confidence interval for \(\mu \).

[3]
b.

Given the hypotheses

\[{{\text{H}}_0}:\mu  = 15;{\text{ }}{{\text{H}}_1}:\mu  \ne 15,\]

find the p-value of the above results and state your conclusion at the 1 % significance level.

[4]
c.
Answer/Explanation

Markscheme

\(\bar x = 14\)     A1

\(s_{n – 1}^2 = \frac{{3977.57}}{{19}} – \frac{{{{280}^2}}}{{380}}\)     (M1)

\( = 3.03\)     A1

[3 marks]

Note: Accept any notation for these estimates including \(\mu \) and \({\sigma ^2}\).

Note: Award M0A0 for division by 20.

a.

the 95% confidence limits are

\(\bar x \pm t\sqrt {\frac{{s_{n – 1}^2}}{n}} \)     (M1)

Note: Award M0 for use of z.

 

ie, \(14 \pm 2.093\sqrt {\frac{{3.03}}{{20}}} \)     (A1)

Note:FT their mean and variance from (a).

 

giving [13.2, 14.8]     A1

Note: Accept any answers which round to 13.2 and 14.8.

 

[3 marks]

b.

Use of t-statistic \(\left( { = \frac{{14 – 15}}{{\sqrt {\frac{{3.03}}{{20}}} }}} \right)\)     (M1)

Note:FT their mean and variance from (a).

 

Note: Award M0 for use of z.

 

Note: Accept \(\frac{{15 – 14}}{{\sqrt {\frac{{3.03}}{{20}}} }}\).

 

\( = – 2.569 \ldots \)     (A1)

Note: Accept \(2.569 \ldots \)

 

\(p{\text{ – value}} = 0.009392 \ldots  \times 2 = 0.0188\)     A1

Note: Accept any answer that rounds to 0.019.

 

Note: Award (M1)(A1)A0 for any answer that rounds to 0.0094.

 

insufficient evidence to reject \({{\text{H}}_0}\) (or equivalent, eg accept \({{\text{H}}_0}\) or reject \({{\text{H}}_1}\))     R1

Note:FT on their p-value.

 

[4 marks]

c.

Examiners report

In (a), most candidates estimated the mean correctly although many candidates failed to obtain a correct unbiased estimate for the variance. The most common error was to divide \(\sum {{x^2}} \) by \(20\) instead of \(19\). For some candidates, this was not a costly error since we followed through their variance into (b) and (c).

a.

In (b) and (c), since the variance was estimated, the confidence interval and test should have been carried out using the t-distribution. It was extremely disappointing to note that many candidates found a Z-interval and used a Z-test and no marks were awarded for doing this. Candidates should be aware that having to estimate the variance is a signpost pointing towards the t-distribution.

b.

In (b) and (c), since the variance was estimated, the confidence interval and test should have been carried out using the t-distribution. It was extremely disappointing to note that many candidates found a Z-interval and used a Z-test and no marks were awarded for doing this. Candidates should be aware that having to estimate the variance is a signpost pointing towards the t-distribution.

c.

Question

The number of machine breakdowns occurring in a day in a certain factory may be assumed to follow a Poisson distribution with mean \(\mu \). The value of \(\mu \) is known, from past experience, to be 1.2. In an attempt to reduce the value of \(\mu \), all the machines are fitted with new control units. To investigate whether or not this reduces the value of \(\mu \), the total number of breakdowns, x, occurring during a 30-day period following the installation of these new units is recorded.

State suitable hypotheses for this investigation.

[1]
a.

It is decided to define the critical region by \(x \leqslant 25\).

(i)     Calculate the significance level.

(ii)     Assuming that the value of \(\mu \) was actually reduced to 0.75, determine the probability of a Type II error.

[8]
b.
Answer/Explanation

Markscheme

\({{\text{H}}_0}:\mu  = 1.2\); \({{\text{H}}_1}:\mu  < 1.2\)     A1

Note: Accept “ \({{\text{H}}_0}:\) (\(30\)-day) mean \( = 36\); \({{\text{H}}_1}:\) (\(30\)-day) mean \( = 36\) ”.

[1 mark]

a.

(i)     let X denote the number of breakdowns in 30 days

then under \({{\text{H}}_0}\) , \(E(X) = 36\)     (A1)

\({\text{sig level}} = {\text{P}}(X \leqslant 25|{\text{mean}} = 36)\)     (M1)(A1)

= 0.0345 (3.45%)     A1

Note: Accept any answer that rounds to 0.035 (3.5%) .

Note: Do not accept the use of a normal approximation.

(ii)     under \({{\text{H}}_1}\), \(E(X) = 22.5\)     (A1)

\(P{\text{(Type II error)}} = P(X \geqslant 26|{\text{mean}} = 22.5)\)     (M1)(A1)
= 0.257     A1

Note: Accept any answer that rounds to 0.26.

Note: Do not accept the use of a normal approximation.

[8 marks]

b.

Examiners report

This question was well answered by many candidates. The most common error was to attempt to use a normal approximation to find approximate probabilities instead of the Poisson distribution to find the exact probabilities. Some candidates appeared not to be familiar with the term ‘Type II error probability’ which made (b)(ii) inaccessible. Another fairly common error was to believe that the complement of \(x \leqslant 25\) is \(x \geqslant 25\).

a.

This question was well answered by many candidates. The most common error was to attempt to use a normal approximation to find approximate probabilities instead of the Poisson distribution to find the exact probabilities. Some candidates appeared not to be familiar with the term ‘Type II error probability’ which made (b)(ii) inaccessible. Another fairly common error was to believe that the complement of \(x \leqslant 25\) is \(x \geqslant 25\).

b.

Question

Jenny tosses seven coins simultaneously and counts the number of tails obtained. She repeats the experiment 750 times. The following frequency table shows her results.

Explain what can be done with this data to decrease the probability of making a type I error.

[2]
b.

(i)     State the meaning of a type II error.

(ii)     Write down how to proceed if it is required to decrease the probability of making both a type I and type II error.

[2]
c.
Answer/Explanation

Markscheme

reduce the significance level (or equivalent statement)     R2

[2 marks]

b.

(i)     accepting \({{\text{H}}_0}\) (or failing to reject \({{\text{H}}_0}\)) when it is false (or equivalent)     A1

(ii)     increase the number of trials     A1

[2 marks]

c.

Examiners report

It was disappointing to see that some candidates wrote incorrect hypotheses, eg ‘\({{\text{H}}_0}\): Data are binomial; \({{\text{H}}_1}\): Data are not binomial’ without specifying any parameters. Part (b) caused unexpected problems for many candidates who misunderstood the question and gave ‘increase the number of trials’ as their answer.

b.

 

c.

Question

Francisco and his friends want to test whether performance in running 400 metres improves if they follow a particular training schedule. The competitors are tested before and after the training schedule.

The times taken to run 400 metres, in seconds, before and after training are shown in the following table.

Apply an appropriate test at the 1% significance level to decide whether the training schedule improves competitors’ times, stating clearly the null and alternative hypotheses. (It may be assumed that the distributions of the times before and after training are normal.)

Answer/Explanation

Markscheme

\({{\text{H}}_0}\): the training schedule does not help improve times (or \(\mu  = 0\))     A1

\({{\text{H}}_1}\): the training schedule does help improve times (or \(\mu  > 0\))     A1

Note: Subsequent marks can be awarded even if the hypotheses are not stated.

     (Assuming difference of times is normally distributed.)

let \(d{\text{ time before training }}-{\text{ time after training}}\)     (M1)



 

EITHER

\(n = 5,{\text{ }}\sum {d = 13,{\text{ }}} \sum {{d^2} = 79 \Rightarrow s_{n – 1}^2 = \frac{1}{4}\left( {79 – \frac{{169}}{5}} \right) = 11.3} \)     (M1)

(small sample) so use a one-sided t-test     (M1)

Note:     The “one-sided” t-test may have been seen above when stating \({{\text{H}}_1}\).

\(t = \frac{{2.6}}{{\sqrt {\frac{{11.3}}{5}} }} = 1.7 \ldots \)     A1

\(v = 4\),     A1

at the 1% level the critical value is 3.7     A1

since \({\text{3.7}} > {\text{1.7}} \ldots \)

\({{\text{H}}_0}\) is accepted (insufficient evidence to reject \({{\text{H}}_0}\))     R1

Note: Follow through their t-value.

OR

(small sample) so use a one-sided t-test     (M1)

\(p = 0.079 \ldots \)     A4

since \(0.079 \ldots  > 0.01\)

\({{\text{H}}_0}\) is accepted (insufficient evidence to reject \({{\text{H}}_0}\))     R1

Note: Follow through their p-value.

Note: Accept \(d = {\text{time after training }}-{\text{ time before training throughout}}\).

[10 marks]

Examiners report

It was again disappointing to see many candidates giving incorrect hypotheses. A common error was to give the hypotheses the wrong way around. Candidates should be aware that in this type of problem the null hypothesis always represents the status quo. Also, some candidates defined ‘\(d = {\text{time before }}-{\text{ time after}}\)’ and then gave the hypotheses incorrectly as \({{\text{H}}_0}:d = 0\) or \(\bar d = 0;{\text{ }}{{\text{H}}_1}:d > 0\) or \(\bar d > 0\). It is important to note that the parameter being tested here is \(E(d)\) or \({\mu _d}\) although \(\mu \) was accepted.

Question

Eleven students who had under-performed in a philosophy practice examination were given extra tuition before their final examination. The differences between their final examination marks and their practice examination marks were

\[10,{\text{ }} – 1,{\text{ }}6,{\text{ }}7,{\text{ }} – 5,{\text{ }} – 5,{\text{ }}2,{\text{ }} – 3,{\text{ }}8,{\text{ }}9,{\text{ }} – 2.\]

Assume that these differences form a random sample from a normal distribution with mean \(\mu \) and variance \({\sigma ^2}\).

Determine unbiased estimates of \(\mu \) and \({\sigma ^2}\).

[4]
a.

(i)     State suitable hypotheses to test the claim that extra tuition improves examination marks.

(ii)     Calculate the \(p\)-value of the sample.

(iii)     Determine whether or not the above claim is supported at the \(5\% \) significance level.

[8]
b.
Answer/Explanation

Markscheme

unbiased estimate of \(\mu \) is \(2.36(36 \ldots )\;\;\;(26/11)\)     (M1)A1

unbiased estimate of \({\sigma ^2}\) is \(33.65(45 \ldots ) = ({5.801^2})\;\;\;(1851/55)\)     (M1)A1

Note:     Accept any answer that rounds correctly to \(3\) significant figures.

Note:     Award M1A0 for any unbiased estimate of \({\sigma ^2}\) that rounds to \(5.80\).

[4 marks]

a.

(i)     \({{\text{H}}_0}:\mu  = 0;{\text{ }}{{\text{H}}_1}:\mu  > 0\)     A1A1

Note:     Award A1A0 if an inappropriate symbol is used for the mean, eg, \(r\), \({\rm{\bar d}}\).

(ii)     attempt to use t-test     (M1)

\(t = 1.35\)     (A1)

\({\text{DF}} = 10\)     (A1)

\(p\)-value \( = 0.103\)     A1

Note:     Accept any answer that rounds correctly to \(3\) significant figures.

(iii)     \(0.103 > 0.05\)     A1

there is insufficient evidence at the \(5\% \) level to support the claim (that extra tuition improves examination marks)

OR

the claim (that extra tuition improves examination marks) is not supported at the \(5\% \) level (or equivalent statement)     R1

Note:     Follow through the candidate’s \(p\)-value.

Note:     Do not award R1 for Accept \({{\text{H}}_0}\) or Reject \({{\text{H}}_1}\).

[8 marks]

Total [12 marks]

b.

Examiners report

Almost every candidate gave the correct estimate of the mean but some chose the wrong variance from their calculators to estimate \({\sigma ^2}\). In (b)(i), the hypotheses were sometimes incorrectly written, usually with an incorrect symbol instead of \(\mu \), for example \(d\), \(\bar x\) and ‘mean’ were seen. Many candidates failed to make efficient use of their calculators in (b)(ii). The intention of the question was that candidates should simply input the data into their calculators and use the software to give the p-value. Instead, many candidates found the p-value by first evaluating \(t\) using the appropriate formula. This was a time consuming process and it gave opportunity for error. In (b)(iii), candidates were expected to refer to the claim so that the answers ‘Accept \({H_0}\)’ or ‘Reject \({H_1}\)’ were not accepted.

a.

Almost every candidate gave the correct estimate of the mean but some chose the wrong variance from their calculators to estimate \({\sigma ^2}\). In (b)(i), the hypotheses were sometimes incorrectly written, usually with an incorrect symbol instead of \(\mu \), for example \(d\), \(\bar x\) and ‘mean’ were seen. Many candidates failed to make efficient use of their calculators in (b)(ii). The intention of the question was that candidates should simply input the data into their calculators and use the software to give the \(p\)-value. Instead, many candidates found the \(p\)-value by first evaluating \(t\) using the appropriate formula. This was a time consuming process and it gave opportunity for error. In (b)(iii), candidates were expected to refer to the claim so that the answers ‘Accept \({H_0}\)’ or ‘Reject \({H_1}\)’ were not accepted.

b.

Question

A smartphone’s battery life is defined as the number of hours a fully charged battery can be used before the smartphone stops working. A company claims that the battery life of a model of smartphone is, on average, 9.5 hours. To test this claim, an experiment is conducted on a random sample of 20 smartphones of this model. For each smartphone, the battery life, \(b\) hours, is measured and the sample mean, \({\bar b}\), calculated. It can be assumed the battery lives are normally distributed with standard deviation 0.4 hours.

It is then found that this model of smartphone has an average battery life of 9.8 hours.

State suitable hypotheses for a two-tailed test.

[1]
a.

Find the critical region for testing \({\bar b}\) at the 5 % significance level.

[4]
b.

Find the probability of making a Type II error.

[3]
c.

Another model of smartphone whose battery life may be assumed to be normally distributed with mean μ hours and standard deviation 1.2 hours is tested. A researcher measures the battery life of six of these smartphones and calculates a confidence interval of [10.2, 11.4] for μ.

Calculate the confidence level of this interval.

[4]
d.
Answer/Explanation

Markscheme

Note: In question 3, accept answers that round correctly to 2 significant figures.

\({{\text{H}}_0}\,{\text{:}}\,\mu  = 9.5{\text{;}}\,\,{{\text{H}}_1}\,{\text{:}}\,\mu  \ne 9.5\)     A1

[1 mark]

a.

Note: In question 3, accept answers that round correctly to 2 significant figures.

the critical values are \(9.5 \pm 1.95996 \ldots  \times \frac{{0.4}}{{\sqrt {20} }}\)     (M1)(A1)

i.e. 9.3247…, 9.6753…

the critical region is \({\bar b}\) < 9.32, \({\bar b}\) > 9.68     A1A1

Note: Award A1 for correct inequalities, A1 for correct values.

Note: Award M0 if t-distribution used, note that t(19)97.5 = 2.093 …

[4 marks]

b.

Note: In question 3, accept answers that round correctly to 2 significant figures.

\(\bar B \sim {\text{N}}\left( {9.8,\,{{\left( {\frac{{0.4}}{{\sqrt {20} }}} \right)}^2}} \right)\)     (A1)

\({\text{P}}\left( {9.3247 \ldots  < \bar B < 9.6753 \ldots } \right)\)     (M1)

=0.0816     A1

Note: FT the critical values from (b). Note that critical values of 9.32 and 9.68 give 0.0899.

[3 marks]

c.

Note: In question 3, accept answers that round correctly to 2 significant figures.

METHOD 1

\(X \sim {\text{N}}\left( {{\text{10}}{\text{.8,}}\,\frac{{{{1.2}^2}}}{6}} \right)\)     (M1)(A1)

P(10.2 < X < 11.4) = 0.7793…     (A1)

confidence level is 77.9%    A1

Note: Accept 78%.

METHOD 2

\(11.4 – 10.2 = 2z \times \frac{{1.2}}{{\sqrt 6 }}\)      (M1)

\(z = 1.224 \ldots \)     (A1)

P(−1.224… < Z < 1.224…) = 0.7793…      (A1)

confidence level is 77.9%      A1

Note: Accept 78%.

[4 marks]

d.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

Question

A shopper buys 12 apples from a market stall and weighs them with the following results (in grams).

117, 124, 129, 118, 124, 116, 121, 126, 118, 121, 122, 129

You may assume that this is a random sample from a normal distribution with mean \(\mu \) and variance \({\sigma ^2}\).

Determine unbiased estimates of \(\mu \) and \({\sigma ^2}\).

[3]
a.

Determine a 99 % confidence interval for \(\mu \) .

[2]
b.

The stallholder claims that the mean weight of apples is 125 grams but the shopper claims that the mean is less than this.

(i)     State suitable hypotheses for testing these claims.

(ii)     Calculate the p-value of the above sample.

(iii)     Giving a reason, state which claim is supported by your p-value using a 5 % significance level.

[5]
c.
Answer/Explanation

Markscheme

unbiased estimate of \(\mu = 122\)     A1

unbiased estimate of \({\sigma ^2} = 4.4406{ \ldots ^2} = 19.7\)     (M1)A1

Note: Award (M1)A0 for 4.44.

 

[3 marks]

a.

the 99 % confidence interval for \(\mu \) is [118, 126]     A1A1

[2 marks]

b.

(i)     \({{\text{H}}_0}:\mu  = 125;{\text{ }}{{\text{H}}_1}:\mu < 125\)     A1

 

(ii)     p-value = 0.0220     A2

 

(iii)     the shopper’s claim is supported because \(0.0220 < 0.05\)     A1R1

[5 marks]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

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