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Question (lecture 17).

In January 2011 Durham police were reported as disappointed by the increase in the num‐ ber of people arrested for drinking and driving. Between December 1st 2010 and December 31st 2010 they had 52 positive breath tests out of 1799 breath tests administered, while for the same period in 2009 they had 41 positive tests out of 1433 administered. Construct a 95% confidence interval for the difference in proportion of drivers who tested positive. Source: http://www.bbc. co.uk/news/uk‐england‐12261462

Answer:

36

Question (lecture 17).

I observe two dice. For one die I notice that it gives a six 20 times out of 100 and for the second die I notice that it gives a six 22 times out of 80. Test at the 5% level whether the two dice give the same37p robability of showing a six.

Answer:

Question (lecture 18).

If Xχ4,382 for what value of x is pr{X>x}=0.05? Answer:

Question (lecture 19).

I roll a die 100 times and observe the following results.

Outcome i 1 2 3 4 5 6 Observed frequency 16 15 16 15 15 23

Test at the 5% level whether the die is fair. Answer:

39

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Two binomial proportions here. θˆ1=4/7=0.571, θˆ2=3/10=0.300, n1=7, n2=10. Common estimated proportion is θ=7θˆ1+10θˆ217=0.412. Approximate test statistic is z=|θˆ1-θˆ2|θˆ(1-θˆ)(1n1+1n2)=1.119. reject Ho at 5% level if |z|>1.96, so here accept the hypothesis that the two proportions are equal.

36 Two binomial proportions again. θˆ1=52/1799=0.028905, θˆ2=41/1433=0.028611, n1=1799, n2=1433. Common estimated proportion is θ=1799θˆ1+1433θˆ23232=0.0288. (This is very small so the normal approxima‐ tion is doubtful. In practice we would transform to give approximate normality.) Approximate test statistic is z=|θˆ1-θˆ2|θˆ(1-θˆ)(1n1+1n2)=0.0496. Reject Ho at 5% level if |z|>1.96, so here accept the hypothesis that the two proportions are equal.

37

n1=100, x1 =20, θˆ1=20/100=0.200, n2 =80, x2 =22, θˆ2=22/80=0.275. We test Ho: θ1= θ2(=θ)vs. H1:θ1θ2. This is equivalent to testing Ho: θ1-θ2=0vs. H1:θ1-θ20. Assuming Ho is true, the estimated common proportion θ is estimated by θˆ=n1θˆ1+n2θˆ2n1+n2=20+22180=0.2333. Test statistic is z=θˆ1-θˆ2θˆ(1-θˆ)n1+θˆ(1-θˆ)n2=0.200-0.27500017889+00014907=-1.31. Test rule is reject Ho if |z|>1.96, so accept Ho at 5% level.

38From tables, x=9.488.

39 Let X denote the outcome of the die. We test whether pr{X=i}=1/6 for all i. Expected frequency for any outcome would then be 100×16=16.667.

Outcome i 1 2 3 4 5 6

Observed frequency Oi 16 15 16 15 15 23

Expected frequency Ei 16.67 16.67 16.67 16.67 16.67 16.67

(Oi-Ei)2/Ei˙ 0.0267 0.1667 0.0267 0.1667 0.1667 2.407 sum =2.960

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