Sampling Distribution of Differences Between Means
1 = 2 1 2 Null Research Sample 1 Sample 2 Population
μ 1 - μ 2 = 0 μ 1 = μ 2 PopulationSampling Distribution
Hypothesis: Women earned higher grades than men Choose a statistic: Mean grade point average (GPA) H 1 : μ F –μ M > 0 H 0 : μ F –μ M = 0 α<.05; one tail σ 2 =.10 n F = 25; n M = 20; ;
μ F –μ M = 0 z = 1.64, <.05 Do not reject Null
CI 95 : Interval estimate: Point estimate: 0.05 ± 1.96 * 0.09 = ,
Hypothesis: GRE Verbal of English speaking students is better than that of Chinese students Choose a statistic: Mean GRE verbal scores H 0 : μ E –μ C = 0 H 1 : μ E –μ C > 0 α<.01; one tail σ = 100 n E = 100; n C = 400; ;
μ E –μ C = 0 z =2.33, <.01 Reject Null
Point estimate: Interval estimate: CI 90 : 40 ± 1.64 * = 21.66,
t as the sampling distribution of means when 2 is not known and sample variance, s 2, is used to estimate 2. Now you need to estimate the mean but also the variance of the population
Variance known; z test Variance unknown, t test
df = 2 df = 9 df = t crit. = -4.30t crit. = 4.30 t crit. = -2.26t crit. = 2.26 z crit. = -1.96z crit. = 1.96 t crit. = 2.04t crit. = Normal
Hypothesis: Word acquisition of the two groups of pre school children are different Choose a statistics: Mean word acquisition test scores H 0 : μ E = μ C H 1 : μ E ≠μ C α<.05; two tails s 2 E =5.286; s 2 C =3.671 n E = 15; n C = 20; ; df = n E – 1 + n C – 1 = 15 – – 1 = 33
μ E –μ C = 0
t.05 (33)= ±2.04 μ E –μ C = 0 Reject Null
Point estimate: Interval estimate: CI 95 : 2.75 ± 2.04 * 0.71 = 1.30,
Independent vs. Dependent or Paired t-tests PersonMid 1M90 2M65 3M78 4F86 5F77 6F89 7F93 df = n M – 1 n F – 1 df = n – 1 Final D
H 0 : μ 1 –μ 2 = 0 H 1 : μ 1 –μ 2 ≠ 0 <0.01/2 2 = 120 n 1 = 12; n 2 = 20; Answers to Homework 3:
μ 1 –μ 2 = 0 < 0.01/2 <.005 z critical = 2.58
.005 z = -2.58z = 2.58 CI 99 = 1 - 2 = 0 not inside the interval estimates
1 - 2 = 0 0.01/2 t critical (30)= 2.75 .005 Do not reject Null. t = 2.6
.005 t = 2.75t = 1 - 2 = 0 falls inside the interval estimates CI 99 =