7 Do Q21.27 from Moore et al, pp 508-9(6 ed: O19.25, p485) Perform a test of sig

Question

7 Do Q21.27 from Moore et al, pp 508-9(6 ed: O19.25, p485) Perform a test of significance at -0.10 to decide ifw omen talk more than men. Use STUDY2 data onlySample standard deviation, DY 2 data gi Sample mcan Sample size, Show ALL steps in the test. EXTRA: USE 90% confidence interval to determine the difference between men and women in the number of word spoken per day: Conservative) df- so (90%) t* = 90% Confidence interval on the difference is. Conclusion IMPORTANT NOTE: What do you notice about the conclusion from these two approaches? NB Confidence Intervals are always 2-sided procedures. .. so a 90% CI and a ten at -0.10 are only identical always IF the test of significance is also 2-sided. If the test is ONE-sided at = 0.10 as above, then it may give a different conclusion (especially if borderline!) and really corresponds to an 80% (0.10 in each side) confidence interval. MARK: Week 11 Page | 6

Explanation / Answer

(a) H0 : Males speak equal words per day as women speak words per day. m = f

Ha : Males speak less words per day as compared to women speask words per day. m < f

(b) test - statistic

Pooled standard deviation sp = sqrt [(nm -1)s2m  + (nf -1)sf2 )/ (nf + nm -2)]

= sqrt [(19 * 83432 + 26 * 79142 )/ 45 ] = 8098

t = (xw - xm)/ sp sqrt [1/nm + 1/nw] = (16569 - 12867)/ 8098 * sqrt (1/27 + 1/20)

t = 3702 / 2389 = 1.55

(c) Degree of freedom for Option 2 = 27 + 20 - 2 = 45

(d) tcr = t45, 0.05 = 2.014

here t < tcr so we shalln't reject the null hypothesis.

P - value is in the range of 0.15 to 0.05 around 0.10

(e) We can conclude that there are no difference in man and women words per minute speak per day.

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