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.

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.

(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]

(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]

(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) 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.