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1. Past experience indicates that the variance, ¦Ò2x, of the scores, X, obtained on the verbal portion of a test is 5,625. Similarly, the variance ¦Ò2y, of the scores, Y, on the mathematical portion is 2,500.
(a) A random sample of size n = 64 yields a sample mean verbal score of 600. Find a 95% confidence interval on the true average score obtained on the verbal aptitude test.
(b) A random sample of size n = 80 yields a sample mean mathematics score of 500. Find a 95% confidence interval on the true average score obtained on the mathematical aptitude test.
(c) Find a 90% confidence interval on ¦Ìy based on the data from (b). Compare the two (95% and 90%) intervals obtained. What does this result imply about the relationship between interval length and degree of confidence?
(d) Ten years prior to the year in which the above data were obtained, ¦Ìy was found to be 520. Can we safely conclude that the average score has declined over the past 10 years? Explain your answer briefly, on the basis of the confidence intervals constructed in (b) and (c)

  • statistic -

    (a) and (b) Variance = SD^2

    95% = mean ± 1.96 SEm

    SEm = SD/√(n-1)

    (c) 90% = mean ± 1.645 SEm

    (d) Z = (mean1 - mean2)/standard error (SE) of difference between means

    SEdiff = √(SEmean1^2 + SEmean2^2)

    SEm = SD/√(n-1)

    If only one SD is provided, you can use just that to determine SEdiff.

    Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.

    I leave the calculations to you.

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