Purchase Solution

# Past exam paper

Not what you're looking for?

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.

##### Solution Preview

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 ...

##### Measures of Central Tendency

Tests knowledge of the three main measures of central tendency, including some simple calculation questions.

##### Terms and Definitions for Statistics

This quiz covers basic terms and definitions of statistics.

##### Measures of Central Tendency

This quiz evaluates the students understanding of the measures of central tendency seen in statistics. This quiz is specifically designed to incorporate the measures of central tendency as they relate to psychological research.