| Date | May 2018 | Marks available | 1 | Reference code | 18M.3sp.hl.TZ0.3 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
| Command term | State | Question number | 3 | Adapted from | N/A |
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.
Find the critical region for testing \({\bar b}\) at the 5 % significance level.
Find the probability of making a Type II error.
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.
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]
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]
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]
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]
