IB DP Further Mathematics -3.6 Testing hypotheses for the mean of a normal population. HL Paper 1

 

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Question

All members of a large athletics club take part in an annual shotput competition.

The following data give the distances achieved, in metres, by a random selection of 10 members of the club in the 2016 competition

11.8, 14.3, 13.8, 10.3, 14.9, 14.7, 12.4, 13.9, 14.0, 11.7

The president of the club wishes to test whether these data provide evidence that distances achieved have increased since the 2015 competition, when the mean result for the club was 12.4 m. You may assume that the distances achieved follow a normal distribution with mean \(\mu \), variance \({\sigma ^2}\), and that the membership of the club has not changed from 2015 to 2016.

State suitable hypotheses.

[1]
a.

(i)     Give a reason why a \(t\) test is appropriate and write down its degrees of freedom.

(ii)     Find the critical region for testing at each of the 5% and 10% significance levels.

[4]
b.

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

(ii)     Find the value of the test statistic.

[3]
c.

State the conclusions that the president of the club should reach from this test, giving reasons for your answer.

[2]
d.
Answer/Explanation

Markscheme

\({H_0}:{\text{ }}\mu  = 12.4;{\text{ }}{H_1}:{\text{ }}\mu  > 12.4\)    A1

[1 mark]

a.

(i)     \(t\) test is appropriate because the variance (standard deviation) is unknown     R1

\(v = 9\)    A1

(ii)     \(t \geqslant 1.83{\text{ }}(5\% );{\text{ }}t \geqslant 1.38{\text{ }}(10\% )\)     A1A1

Note:     Accept strict inequalities.

[4 marks]

b.

(i)     unbiased estimate of \(\mu \) is 13.18     A1

Note:     Accept 13.2.

unbiased estimate of \({\sigma ^2}\) is 2.34 \(({1.531^2})\)     A1

(ii)     \({t_{{\text{calc}}}} = \left( {\frac{{13.18 – 12.4}}{{\frac{{1.531}}{{\sqrt {10} }}}}} \right) = 1.61{\text{ or }}1.65\)     A1

[3 marks]

c.

as \(1.38 < 1.61 < 1.83\)     R1

evidence to accept \({H_0}\) at the 5% level, but not at the 10% level     A1

Note:     Accept the use of the \(p\)value \( = 0.0708\).

[2 marks]

d.

Question

A sample of size 100 is taken from a normal population with unknown mean μ and known variance 36.

Another investigator decides to use the same data to test the hypotheses H0 : μ = 65 , H1 : μ = 67.9.

An investigator wishes to test the hypotheses H0 : μ = 65, H1 : μ > 65.

He decides on the following acceptance criteria:

Accept H0 if the sample mean \(\bar x\) ≤ 66.5

Accept H1 if \(\bar x\) > 66.5

Find the probability of a Type I error.

[3]
a.

She decides to use the same acceptance criteria as the previous investigator. Find the probability of a Type II error.

[3]
b.i.

Find the critical value for \({\bar x}\) if she wants the probabilities of a Type I error and a Type II error to be equal.

[3]
b.ii.
Answer/Explanation

Markscheme

\(\bar X \sim {\text{N}}\left( {\mu ,\,\frac{{{\sigma ^2}}}{n}} \right)\)

\(\bar X \sim {\text{N}}\left( {65,\,\frac{{36}}{{100}}} \right)\)     (A1)

P(Type I Error) \( = {\text{P}}\left( {\bar X > 66.5} \right)\)      (M1)

= 0.00621       A1

[3 marks]

a.

P(Type II Error) = P(accept H0 | H1 is true)

\( = {\text{P}}\left( {\bar X \leqslant 66.5\left| {\mu  = 67.9} \right.} \right)\)        (M1)

\( = {\text{P}}\left( {\bar X \leqslant 66.5} \right)\) when \(\bar X \sim {\text{N}}\left( {67.9,\,\frac{{36}}{{100}}} \right)\)        (M1)

= 0.00982      A1

[3 marks]

b.i.

the variances of the distributions given by H0 and H1 are equal,       (R1)

by symmetry the value of \({\bar x}\) lies midway between 65 and 67.9      (M1)

\( \Rightarrow \bar x = \frac{1}{2}\left( {65 + 67.9} \right) = 66.45\)       A1

[3 marks]

b.ii.

Question

At an early stage in analysing the marks scored by candidates in an examination paper, the examining board takes a random sample of 250 candidates and finds that the marks, \(x\) , of these candidates give \(\sum {x = 10985} \) and \(\sum {{x^2} = 598736} \).

Calculate a 90% confidence interval for the population mean mark μ for this paper.

[4]
a.

The null hypothesis μ = 46.5 is tested against the alternative hypothesis μ < 46.5 at the λ% significance level. Determine the set of values of λ for which the null hypothesis is rejected in favour of the alternative hypothesis.

[4]
b.
Answer/Explanation

Markscheme

\(\bar x = 43.94\)      (A1)

unbiased variance estimate = 466.0847        (A1)

Note: Accept sample variance = 464.2204.

⇒ 90% confidence interval is (41.7,46.2)       A1A1

[4 marks]

a.

Z-value is −1.87489 or −1.87866       (A1)

probability is 0.0304 or 0.0301      (A1)

λ ≥ 3.01       (M1)A1

[4 marks]

b.
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