IB DP Maths Topic 7.6 Critical regions, critical values, p-values, one-tailed and two-tailed tests HL Paper 3

 

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Question

A coin was tossed 200 times and 115 of these tosses resulted in ‘heads’. Use a two-tailed test with significance level 1 % to investigate whether or not the coin is biased.

Answer/Explanation

Markscheme

The number of’ ‘heads’ X is B(200, p)     (M1)

\({{\text{H}}_0}:p = 0.5;{\text{ }}{{\text{H}}_1}:p \ne 0.5\)     A1A1

Note: Award A1A0 for the statement “ \({{\text{H}}_0}:\) coin is fair; \({{\text{H}}_1}:\) coin is biased”.

 

EITHER

\({\text{P}}(\left. {X \geqslant 115} \right|{{\text{H}}_0}) = 0.0200\)     (M1)(A1)

p-value = 0.0400     A1

This is greater than 0.01.     R1

There is insufficient evidence to conclude that the coin is biased (or the coin is not biased).     R1

OR

(Using a proportion test on a GDC) p-value = 0.0339     N3

This is greater than 0.01.     R1

There is insufficient evidence to conclude that the coin is biased (or the coin is not biased).     R1

OR

Under \({{\text{H}}_0}X\) is approximately N(100, 50)     (M1)

\(z = \frac{{115 – 100}}{{\sqrt {50} }} = 2.12\)     (M1)A1

(Accept 2.05 with continuity correction)

This is less than 2.58     R1

There is insufficient evidence to conclude that the coin is biased (or the coin is not biased).     R1

OR

99 % confidence limits for p are \(\frac{{115}}{{200}} \pm 2.576\sqrt {\frac{{115}}{{200}} \times \frac{{85}}{{200}} \times \frac{1}{{200}}} \)     (M1)A1

giving [0.485, 0.665]     A1

This interval contains 0.5     R1

There is insufficient evidence to conclude that the coin is biased (or the coin is not biased).     R1

[8 marks]

Examiners report

This question was well answered in general with several correct methods seen. The most popular method was to use a GDC to carry out a proportion test which is equivalent to using a normal approximation. Relatively few candidates calculated an exact p-value using the binomial distribution. Candidates who found a 95% confidence interval for p, the probability of obtaining a head, and noted that this contained 0.5 were given full credit.

Question

A factory makes wine glasses. The manager claims that on average 2 % of the glasses are imperfect. A random sample of 200 glasses is taken and 8 of these are found to be imperfect.

Test the manager’s claim at a 1 % level of significance using a one-tailed test.

Answer/Explanation

Markscheme

Let X denote the number of imperfect glasses in the sample     (M1)

For recognising binomial or proportion or Poisson     A1

(\(X \sim {\text{B}}(200,{\text{ }}p)\) where p-value is the probability of a glass being imperfect)

Let \({{\text{H}}_0}:p{\text{-value}} = 0.02{\text{ and }}{{\text{H}}_1}:p{\text{-value}} > 0.02\)     A1A1

EITHER

p-value = 0.0493     A2

Using the binomial distribution \(p{\text{-value}} = 0.0493 > 0.01{\text{ we accept }}{{\text{H}}_0}\)     R1

OR

p-value = 0.0511     A2

Using the Poisson approximation to the binomial distribution since \(p{\text{-value}} = 0.0511 > 0.01{\text{ we accept }}{{\text{H}}_0}\)     R1

OR

p-value = 0.0217     A2

Using the one proportion z-test since \(p{\text{-value}} = 0.0217 > 0.01{\text{ we accept }}{{\text{H}}_0}\)     R1

Note: Use of critical values is acceptable.

 

[7 marks]

Examiners report

Many candidates used a t-test on this question. This was possibly because the sample was large enough to approximate normality of a proportion. The need to use a one-tailed test was often missed. When using the z-test of proportions p = 0.04 was often used instead of p = 0.02 . Not many candidates used the binomial distribution.

Question

A population is known to have a normal distribution with a variance of 3 and an unknown mean \(\mu \) . It is proposed to test the hypotheses \({{\text{H}}_0}:\mu  = 13,{\text{ }}{{\text{H}}_1}:\mu  > 13\) using the mean of a sample of size 2.

(a)     Find the appropriate critical regions corresponding to a significance level of

  (i)     0.05;

  (ii)     0.01.

(b)     Given that the true population mean is 15.2, calculate the probability of making a Type II error when the level of significance is

  (i)     0.05;

  (ii)     0.01.

(c)     How is the change in the probability of a Type I error related to the change in the probability of a Type II error?

Answer/Explanation

Markscheme

(a)     With \({{\text{H}}_0},{\text{ }}\bar X \sim {\text{N}}\left( {13,\frac{3}{2}} \right) = {\text{N(13, 1.5)}}\)     (M1)(A1)

(i)     5 % for N(0,1) is 1.645

so \(\frac{{\bar x – 13}}{{\sqrt {1.5} }} = 1.645\)     (M1)(A1)

\(\bar x = 13 + 1.645\sqrt {1.5} \)

\( = 15.0\,\,\,\,\,{\text{(3 s.f.)}}\)     A1     N0

\({\text{[15.0, }}\infty {\text{[}}\)

 

(ii)     1% for N(0, 1) is 2.326

so \(\frac{{\bar x – 13}}{{\sqrt {1.5} }} = 2.326\)     (M1)(A1)

\(\bar x = 13 + 2.326\sqrt {1.5} \)

\( = 15.8\,\,\,\,\,{\text{(3 s.f., accept 15.9)}}\)     A1     N0

\({\text{[15.8, }}\infty {\text{[}}\)

[8 marks]

 

(b)     (i)     \(\beta = {\text{P}}(\bar X < 15.0147)\)     M1

\( = 0.440\)     A2

 

(ii)     \(\beta = {\text{P}}(\bar X < 15.8488)\)     M1

\( = 0.702\)     A2

[6 marks]

 

(c)     The probability of a Type II error increases when the probability of a Type I error decreases.     R2

[2 marks]

 

Total [16 marks]

Examiners report

This question proved to be the most difficult. The range of solutions ranged from very good to very poor. Many students thought that \(P(TypeI) = 1 – P(TypeII)\) when in fact \(1 – P(TypeII)\) is the power of the test.

Question

The mean weight of a certain breed of bird is believed to be 2.5 kg. In order to test this belief, it is planned to determine the weights \({x_1}{\text{ , }}{x_2}{\text{ , }}{x_3}{\text{ , }} \ldots {\text{, }}{x_{16}}\) (in kg) of sixteen of these birds and then to calculate the sample mean \({\bar x}\) . You may assume that these weights are a random sample from a normal distribution with standard deviation 0.1 kg.

(a)     State suitable hypotheses for a two-tailed test.

(b)     Find the critical region for \({\bar x}\) having a significance level of 5 %.

(c)     Given that the mean weight of birds of this breed is actually 2.6 kg, find the probability of making a Type II error.

Answer/Explanation

Markscheme

(a)     \({H_0}:\mu = 2.5\)     A1

\({H_1}:\mu \ne 2.5\)     A1

[2 marks]

 

(b)     the critical values are \(2.5 \pm 1.96 \times \frac{{0.1}}{{\sqrt {16} }}\) ,     (M1)(A1)(A1)

i.e. 2.45, 2.55     (A1)

the critical region is \(\bar x < 2.45 \cup \bar x > 2.55\)     A1A1

Note: Accept \( \leqslant ,{\text{ }} \geqslant \) .

 

[6 marks]

 

(c)     \({\bar X}\) is now \({\text{N}}(2.6,{\text{ }}{0.025^2})\)     A1

a Type II error is accepting \({H_0}\) when \({H_1}\) is true     (R1)

thus we require

\({\text{P}}(2.45 < \bar X < 2.55)\)     M1A1

\( = 0.0228\,\,\,\,\,\)(Accept 0.0227)     A1

Note: If critical values of 2.451 and 2.549 are used, accept 0.0207.

 

[5 marks]

Total [13 marks]

Examiners report

In (a), some candidates incorrectly gave the hypotheses in terms of \({\bar x}\) instead of \(\mu \). In (b), many candidates found the correct critical values but then some gave the critical region as \(2.45 < \bar x < 2.55\) instead of \(\bar x < 2.45 \cup \bar x > 2.55\) Many candidates gave the critical values correct to four significant figures and therefore were given an arithmetic penalty. In (c), many candidates correctly defined a Type II error but were unable to calculate the corresponding probability.

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

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

Two species of plant, \(A\) and \(B\), are identical in appearance though it is known that the mean length of leaves from a plant of species \(A\) is \(5.2\) cm, whereas the mean length of leaves from a plant of species \(B\) is \(4.6\) cm. Both lengths can be modelled by normal distributions with standard deviation \(1.2\) cm.

In order to test whether a particular plant is from species \(A\) or species \(B\), \(16\) leaves are collected at random from the plant. The length, \(x\), of each leaf is measured and the mean length evaluated. A one-tailed test of the sample mean, \(\bar X\), is then performed at the \(5\% \) level, with the hypotheses: \({H_0}:\mu  = 5.2\) and \({H_1}:\mu  < 5.2\).

Find the critical region for this test.

[3]
a.

It is now known that in the area in which the plant was found \(90\% \) of all the plants are of species \(A\) and \(10\% \) are of species \(B\).

Find the probability that \(\bar X\) will fall within the critical region of the test.

[2]
c.

If, having done the test, the sample mean is found to lie within the critical region, find the probability that the leaves came from a plant of species \(A\).

[3]
d.
Answer/Explanation

Markscheme

\(\bar X \sim N\left( {5.2,{\text{ }}\frac{{{{1.2}^2}}}{{16}}} \right)\)     (M1)

critical value is \(5.2 – 1.64485 \ldots  \times \frac{{1.2}}{4} = 4.70654 \ldots \)     (A1)

critical region is \(] – \infty ,{\text{ }}4.71]\)     A1

Note:     Allow follow through for the final A1 from their critical value.

Note:     Follow through previous values in (b), (c) and (d).

[3 marks]

a.

\(0.9 \times 0.05 + 0.1 \times (1 – 0.361 \ldots ) = 0.108875997 \ldots  = 0.109\)     M1A1

Note:     Award M1 for a weighted average of probabilities with weights \(0.1,0.9\).

[2 marks]

c.

attempt to use conditional probability formula     M1

\(\frac{{0.9 \times 0.05}}{{0.108875997 \ldots }}\)     (A1)

\( = 0.41334 \ldots  = 0.413\)     A1

[3 marks]

Total [10 marks]

d.

Examiners report

Solutions to this question were generally disappointing.

In (a), the standard error of the mean was often taken to be \(\sigma (1.2)\) instead of \(\frac{\sigma }{{\sqrt n }}(0.3)\) and the solution sometimes ended with the critical value without the critical region being given.

a.

In (c), the question was often misunderstood with candidates finding the weighted mean of the two means, ie \(0.9 \times 5.2 + 0.1 \times 4.6 = 5.14\) instead of the weighted mean of two probabilities.

c.

Without having the solution to (c), part (d) was inaccessible to most of the candidates so that very few correct solutions were seen.

d.

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.

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