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Question (lecture 15).

Two independent samples gave values 3, 6, 5, 2 for sample 1 and 2, 2, 3, 3, 5 for sample 2. Assuming that the samples come from independent normal distributions with common unknown variance σ2, test at the 5% level whether the difference in mean equals zero against the alternative that it does not equal zero.

Answer:

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Question (lecture 15).

Five randomly selected remuneration packages for US oil and gas CEOs in 2008 were (in thousands of US dollars) 21333, 7294, 6712, 5727, 7087. Five randomly selected remuneration packages for US health care CEOs in 2008 were (in thousands of dollars) 14262, 8381, 7245, 10211, 1817. Test at the 5% level whether the difference in mean remuneration equals zero against the alternative hypothesis that it does not equal zero. You can assume that the two populations have common (unknown) variance σ2.

Answer:

32

Question (lecture 16).

A quarter of insurance claims are incomplete in some way. If you have 250 forms to process, what is the approximate probability that you will find fewer than 50 of them incomplete?

Answer:

33

Question (lecture 16).

In n=100 tosses of a coin I obtain X=72 heads. Obtain an approximate 95% confidence interval for the probability θ34 of a head.

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Question (lecture 17).

In December 2010 two analysts suggested several shares as likely to rise in 2011. By the end of October 2011 one (Neil Woodford) had four out of n1=7 ""share tips“ showing a rise while the other (Harry Nummo) had three out of n2=10 ""share tips“ showing a rise. Test at the 5% level whether the two success proportions are significantly different.

Answer:

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z=x¯1-x¯2σ12n1+σ22n2=4-344+15=0.913. Test rule is reject H0 if |z|>1.96. Thus accept H0 at 5% level.

31n1=4, x¯1=4, s12=3.333, n2=5, x¯2=3, s22=1.5, pooled estimate of σ2 is s2=3s12+4s227=2.2857. Testing H0:μ1-μ2=0vs. H1:μ1-μ20. Test statistic is t=x¯1-x¯2s1n1+1n2=4-31.5119×14+15=0.986. Test rule is reject H0 if |t| >t7(2.5%). As t7(2.5%) =2.365, accept H0 at 5% level.

32 Data source: http://graphicsweb.wsj. com/php/CE0PAY09.html.

n1=5, x¯1=9630.6, s12=43158021, n2=5, x¯2=8383.2, s22=20577907, n1+n2-2=8, t8(2.5%) =2.306. If variances are equal to σ2, estimate σ2 using s2=(n1-1)s12+(n2-1)s22n1+n2-2=31867964. Test statistic is t= |x¯1-x¯2|s2(1n1+1n2)=1247.43570.32=0.349. Since t8(2.5%) =2.306, then |t|<t8(2.5%) so accept H0 that μ1=μ2 against the alternative μ1μ2 at the 5% level.

33 If X is the number of incomplete forms, XBin(n=250,θ=14)N(μ=62.5,σ2=46.875) . You require pr{X<50}=pr{X49}=Φ(49+12-μσ)=Φ(-1.899)=0.0288. Notice we have used acontinuity correction. 34 Number of heads XBin(n=100,θ) . Here n=100, X=72 observed, θˆ=X/n=72/100=0.72. Approximate 95% confidence interval is θˆ±1.96θˆ(1-θˆ)n=0.72±0.088.

35 Data source: http://www.thisismoney.co.uk/money/investing/article‐1709914/Stock‐market‐predict

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