# IB DP Maths Topic 7.6 Significance level HL Paper 3

## 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.


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.


b.

## 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

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.


a.

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


b.

Find the probability of making a Type II error.


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.


d.

## 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.

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b.

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c.

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d.