Mark Scheme

[Maximum Mark: 6]
(a) Hypotheses
  • Null: \(H_0:\ \mu_B=\mu_P\) (or \(\mu_B-\mu_P=0\)). (1 mark)
  • Alternative (one-tailed): \(H_1:\ \mu_B>\mu_P\) (Beagles are heavier on average). (1 mark)
(b) p-value using a TI-84 Plus CE
  • Enter data:
    • Press [STAT] → 1:Edit…. Put Beagles in L1: 15.3, 16.1, 14.8, 16.4, 15.6, 16.2, 15.0, 16.0, 15.7, 16.3.
    • Put Poodles in L2: 14.5, 14.7, 14.3, 14.9, 15.1, 14.8, 14.6, 14.7, 14.4, 14.9.
    (Correct lists set up. 0.5 mark)
  • Run two-sample t-test: Press [STAT] → TESTS → 4:2-SampTTest. Choose Data; List1=L1, List2=L2, Freq1=Freq2=1, select \(\mu_1>\mu_2\), set Pooled: No, then Calculate. (Correct test and settings. 0.5 mark)
  • TI-84 output (approx.): \(\bar x_1=15.74,\ s_1=0.558,\ n_1=10;\ \bar x_2=14.69,\ s_2=0.247,\ n_2=10\). Test statistic \(t\approx 5.44\), df \(\approx 12.39\), and one-tailed p-value \(p\approx 6.7\times10^{-5}\). (Using “Pooled: Yes” gives the same \(t\) with df \(=18\) and \(p\approx1.8\times10^{-5}\); either is acceptable.) (1 mark)
(c) Conclusion
  • Since \(p\ll 0.05\) (indeed \(p<0.001\)), reject \(H_0\). (1 mark)
  • Conclusion: There is strong statistical evidence that Beagles are heavier on average than Poodles. (1 mark)
Award full credit for equivalent correct TI-84 procedures and answers close to: \(t\approx5.44\), \(p\) between \(2\times10^{-5}\) and \(7\times10^{-5}\); conclusion: reject \(H_0\) in favor of \(\mu_B>\mu_P\).
Scroll to Top