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Confidence intervals from small samplesAQA A-Level Further Maths: Subtopic test

10 questions, 27 marks

AQA A-Level Further Maths

Confidence intervals from small samples

Total 27 marks

Name

Class

Date

  1. 1
    The masses of chocolate bars from a production line are normally distributed. A random sample of 88 bars has sample mean 36.236.2 g and an unbiased estimate of the population standard deviation of 2.42.4 g.
    (a)
    State the number of degrees of freedom of the tt-distribution used to construct a confidence interval for the population mean.
    [1 mark]
    • A88
    • B66
    • C99
    • D77
    (b)
    The 2.5%2.5\% critical value of tt with 77 degrees of freedom is 2.3652.365. Find a 95%95\% confidence interval for the population mean mass.
    [1 mark]
    • A(34.54, 37.86)(34.54,\ 37.86)
    • B(34.19, 38.21)(34.19,\ 38.21)
    • C(34.24, 38.16)(34.24,\ 38.16)
    • D(30.52, 41.88)(30.52,\ 41.88)
    (c)
    The 0.5%0.5\% critical value of tt with 77 degrees of freedom is 3.4993.499. Find a 99%99\% confidence interval for the population mean mass.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    The battery life, in hours, of a make of phone is normally distributed. A random sample of 1212 phones has ∑x=220.8\sum x=220.8 and ∑x2=4087.47\sum x^2=4087.47.
    (a)
    Find an unbiased estimate of the population variance.
    [1 mark]
    • A2.252.25 hours2^2
    • B2.062.06 hours2^2
    • C1.501.50 hours2^2
    • D0.190.19 hours2^2
    (b)
    The 5%5\% critical value of tt (one tail) with 1111 degrees of freedom is 1.7961.796. Find a 90%90\% confidence interval for the population mean battery life.
    [1 mark]
    • A(17.69, 19.11)(17.69,\ 19.11)
    • B(17.45, 19.35)(17.45,\ 19.35)
    • C(17.62, 19.18)(17.62,\ 19.18)
    • D(15.71, 21.09)(15.71,\ 21.09)
    (c)
    The manufacturer claims that the mean battery life is 2020 hours. Use your interval from (b) to comment on this claim.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    The reaction times, in milliseconds, of a random sample of 77 drivers are 248, 262, 255, 241, 270, 259, 253248,\ 262,\ 255,\ 241,\ 270,\ 259,\ 253. Reaction times may be assumed to be normally distributed.
    (a)
    Calculate the sample mean and an unbiased estimate of the population variance.
    [3 marks]
    (b)
    The 2.5%2.5\% critical value of tt with 66 degrees of freedom is 2.4472.447. Construct a 95%95\% confidence interval for the mean reaction time.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    A chemist measures the percentage purity of aspirin in a random sample of 55 tablets from one batch: 98.1, 97.4, 98.6, 97.9, 98.098.1,\ 97.4,\ 98.6,\ 97.9,\ 98.0. Purities are normally distributed.
    (a)
    (i) The 2.5%2.5\% critical value of tt with 44 degrees of freedom is 2.7762.776. Construct a 95%95\% confidence interval for the mean purity.
    (ii) The manufacturer claims that the mean purity is
    98.5%98.5\%. Comment on this claim using your interval.
    (iii) Explain why a
    tt-distribution is used rather than a normal distribution.
    [6 marks]
    (b)
    The chemist then takes a random sample of 2525 tablets from a second batch. The sample mean is 97.98%97.98\% and the unbiased estimate of the population standard deviation is 0.52%0.52\%.
    (i) The
    2.5%2.5\% critical value of tt with 2424 degrees of freedom is 2.0642.064. Construct a 95%95\% confidence interval for the mean purity of the second batch.
    (ii) Compare the width of this interval with that in (a)(i), and evaluate what the comparison shows.
    [6 marks]

    Total for question 4: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).