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Past exam paper

I have a very congested exam timetable and have limited time to revise as i have 4 in a row. if anyone could give me the solutions and workings to the attatched exam paper i would be very greatful.

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A1 Refer to the EXCEL file for calculation:
<br>Mean=SUM(x*Prob(X=x)) = 2.25
<br>Var = SUM((x-Mean)^2 *Prob(X=x)) = 6.80
<br>Standard Deviation = SQRT(Var) = 2.61
<br>Standard Error = SD / SQRT(N) = 2.61 / SQRT(20) = 0.58
<br>Calculate z value by
<br>Z = (2.5-M)/ SE = (2.5-2.25)/0.58= 0.429
<br>By a z table, Prob(z&lt;0.429) = 0.666
<br>Then Prob (Xmean &gt; 2.5) = Prob(z&gt;0.429)= 1 - Prob (z&lt;0.429) = 1-0.666= 0.334 = 33.4%
<br>
<br>A2. Confidence Interval
<br>a) A confidence interval is a range or interval of numbers believed to include an unknown population parameter. Associated with the interval is a measure of the confidence we have that the interval does indeed contain the parameter of interest. If independent samples are taken repeatedly from the same population, and a confidence interval calculated for each sample, then a certain percentage ( confidence level) of the intervals will include the unknown population parameter . Confidence intervals are usually calculated so that this percentage is 95%, but we can produce 90%, 99%, confidence intervals for the unknown parameter.
<br>Because the estimated range of value given by confidence interval is calculated from a given set of sample data. The width of the confidence interval gives us some idea about how uncertain we are about the unknown parameter (see precision). A very wide interval may indicate that more data should be collected before anything very definite can be said about the parameter. Therefore, an "interval estimate" is sometimes thought more appropriate.
<br>b) n=15, Xm=14.8, s=4.2
<br>Standard Error = s / SQRT(N) = 4.2 / SQRT(15) = 1.084
<br>This is a two-tailed t distribution, so we use 99.5% t value with df=15-1=14
<br>by a t table, t (99.5%, 14) = 1.761
<br>Then CI = Xm + - t*SE = 14.8 + - 1.761*1.084 = (12.89, 16.71)
<br>Since the 95% t value with df=14 is 0.692, smaller than the 99% t value, the 95% confidence interval will be narrower than the 99% CI.
<br>
<br>ii) we need to find the CI of population variance first:
<br>by formula, the lower bound is
<br>(n-1)s2 / X2 (n-1) right or,
<br>(n-1)s^2 / Chi-Square(df=n-1) right
<br>from Chi-Square table, since we need 5% on either side, we find
<br>Chi-Square(95%, 14)= 23.68
<br>= 14 * 4.2^2 / 23.68 = 10.43
<br>Then the lower bound of 90% CI of the population Standard Deviation
<br>= SQRT(10.43) = 3.23
<br>
<br>A3. An F-test is used to test if the standard deviations of two populations are equal. This test can be a two-tailed test or a one-tailed test. The two-tailed version tests against the alternative that the standard deviations are not equal.
<br>H0: s1 = s2
<br>Ha: s1 &#8800; s2
<br>Test Statistic: F = (s2 / s1)^2= (6.8 / 3.6)^2 = 3.568
<br>The more this ratio deviates from 1, the stronger the evidence for unequal population variances.
<br>From a F table, the 95% critical F value with ...

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