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t tests for a mean and the paired t-testEdexcel A-Level Further Maths: Flashcards

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When is a $t$ test used instead of a $z$ test for a mean?

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When is a tt test used instead of a zz test for a mean?
When the population variance is unknown and estimated by s2s^2 from the sample, with a Normal population.
Test statistic for a one-sample tt test?
t=xˉ−μ0s/nt=\frac{\bar x-\mu_0}{s/\sqrt n}
Degrees of freedom for a one-sample tt test?
n−1n-1
How does the tt distribution compare with the Normal distribution?
Symmetric but with heavier tails; it approaches N(0,1)\mathrm{N}(0,1) as nn increases.
Assumption for a one-sample tt test?
The population is Normally distributed.
Confidence interval for a mean with σ\sigma unknown?
xˉ±tn−1sn\bar x\pm t_{n-1}\frac{s}{\sqrt n}
Which tt point gives a 95% two-tailed interval?
The upper 2.5%2.5\% point of tn−1t_{n-1}.
Why is a tt interval wider than a zz interval?
It allows for the extra uncertainty from estimating σ\sigma by ss.
When is a paired test used?
When observations come in linked pairs, e.g. the same subjects before and after.
What is the first step in a paired tt-test?
Calculate the difference for each pair.
Test statistic for the paired tt-test?
t=dˉsd/nt=\frac{\bar d}{s_d/\sqrt n} with n−1n-1 degrees of freedom, nn the number of pairs.
What is assumed Normal in the paired test?
The population of differences.
Why not use a two-sample test on paired data?
The samples are not independent; a paired test removes variation between individuals.
What should a conclusion say?
In context, as evidence at a stated level, e.g. 'there is evidence at the 5% level that the mean ...'.

Exam questions on t tests for a mean and the paired t-test

  1. The mass of a chocolate bar is Normally distributed. The label states a mean mass of 5050 g. A random sample of 1010 bars has sample mean 49.249.2 g and sample standard deviation s=2.1s=2.1 g. A test is carried out to see whether the mean mass is less than the label states.
    The lower 5%5\% point of t9t_9 is −1.833-1.833. Complete the test at the 5%5\% significance level and state your conclusion in context.2 marks
  2. A random sample of 88 observations from a Normal population has sample mean 24.324.3 and sample standard deviation s=1.9s=1.9. The population variance is unknown.
    Explain why the tt distribution is used rather than the Normal distribution, and state the assumption needed about the population.2 marks
  3. A supplier claims that the mean length of its rods is 12.012.0 cm. The lengths are Normally distributed. A random sample of 88 rods has ∑x=99.4\sum x=99.4 and ∑x2=1236.48\sum x^2=1236.48, where xx is the length in cm.
    Calculate the sample mean and an unbiased estimate of the population variance.3 marks
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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).