90% 5% 2 5% 1 Module 7 Exercises 1. Several species of bird show hardly any sexual dimorphism; that is, female and male speci- mens look very similar. One such species is Fisher's love bird - a species of small parrot that is native to Africa. To see if there is any difference in the mean height between adult male and female love birds, researchers measured the heights of 23 male and 19 female love birds (ran- domly selected), giving a sample mean of 15.83 cm with a sample standard deviation of 2.1 cm for the male birds and a sample mean of 14.55 cm with a sample standard deviation of 1.9 cm for the female birds. (a) Give an approximate 90% confidence interval for the difference in mean height between male and female love birds. males nm = 23 5m= 15.83 Sm = 2.1 females nf = 19 If = 14.55 Sf=1.9 CI ( 5Cm - 5cf) + +of s.e.(5cm-5g) s . e . ( 5 m - xCf ) - - Xxx ) - 1 Sm + 5 nm = 2.12 + 1.92 = 0.6178 23 19 conservative of = min ( nm - 1 , nf -1 ) =min(22, 18)=18 For a 90% CI we need the 0.95 quantile of the tig-distribution t* = 1.734 90% CI (15.83-14.55) = 1.734 × 0.6178 1.28 # 1.0171 cm (b) State the null and alternative hypotheses that the researchers want to test. Let um be the population mean height of makes " Mf females. Ho: Um = Mg H1 : Mm + Mf
(c) Help the researchers answer their question by carrying out an appropriate hypothesis test. Write your conclusion in a manner that can be understood by the researchers. test statistic t = (5cm-x(+) -(um-Mf)<under s. e. (Jim - xCf ) Ho = (15.83-14.55) - 0 = 2.0717 0.6178 Under Ho, the test statistic has (approximately) a t18 -distribution (again conservative of from (a) P-value = 2xP (T18 > 2.0717) = 2×0.02647 = 0.0529 There is weak evidence against Ho, -2.0717 2-0717 suggesting a difference in mean heights of male + female love birds. 2. Suppose we toss a coin 15 times. What is the smallest number of heads or tails that would give evidence that the coin is not fair at the 5% significance level? Justify your answer.
5. Researchers conducted an experiment to see whether sleep deprivation affected the internal clock. They recruited 16 subjects for one night and randomly assigned them to either a rested group or a sleep-deprived group. For the latter group a research assistant remained with the participants throughout the night to ensure they were awake. Access to television and games were provided. No food was allowed after midnight and caffeinated beverages were discontinued for 24 hours prior to the study. In the morning all participants completed a duration-production task where they were asked to press and hold a button for 1100 ms. The mean recorded duration for the 8 rested subjects is