Question

The following Exercise is based on summary statistics rather than raw data. This information is typically all that is presented in published reports. You can calculate inference procedures by hand from the summaries. Use the conservative Option 2 (degrees of freedom the smaller of n1−1 and n2−1 for two-sample t confidence intervals and P-values. You must trust that the authors understood the conditions for inference and verified that they apply. This isn't always true.) Equip male and female students with a small device that secretly records sound for a random 30 seconds during each 12.5-minute period over two days. Count the words each subject speaks during each recording period, and from this, estimate how many words per day each subject speaks. The published report includes a table summarizing six such studies. Here are two of the six: Sample Size Estimated Average Number (SD) of Words Spoken per Day Study Women Men Women Men 1 51 51 15585 (7449) 16189 (9313) 2 25 22 16601 (7909) 12838 (8188) Readers are expected to understand this to mean, for example, the 51 women in the first study had x⎯⎯⎯ = 15585 and s = 7449. It is commonly thought that women talk more than men. Does either of the two samples support this idea? For each study: (a) State the alternative hypothesis in terms of the population mean number of words spoken per day for men (μM) and for women (μF). If necessary, use != to represent ≠. Study 1: μM < μF Study 2: μM < μF (b) Find the two-sample t statistic to test μF−μM. Study 1: Study 2: (c) What degrees of freedom does Option 2 use to get a conservative P-value? Study 1: 101 Study 2: 46 (d) Compare your value of t with the critical values in Table C. What can you say about the P- value of the test? Study 1: P- value > Study 2: ≤ P- value ≤

Answer 1

Study 1:

For Women:

x̅1 = 15585, s1 = 7449, n1 = 51

For Men:

x̅2 = 16189, s2 = 9313, n2 = 51

Study 2:

For Women:

x̅1 = 16601, s1 = 7909, n1 = 25

For Men:

x̅2 = 12838, s2 = 8188, n2 = 22

a) Alternative hypothesis:

Study 1: µM < µF

Study 2: µM < µF

b) Test statistic:

Study 1:

t = (x̅1 - x̅2)/√(s1²/n1 + s2²/n2) = (15585 - 16189)/√(7449²/51 + 9313²/51) = -0.36169

Study 2:

t = (x̅1 - x̅2)/√(s1²/n1 + s2²/n2) = (16601 - 12838)/√(7909²/25 + 8188²/22) = 1.59737

c)

Study 1: Degree of freedom = min(51-1, 51-1) = 50

Study 2: Degree of freedom = min(25-1, 22-1) = 21

d) P-value:

Study 1: p-value = T.DIST.RT(-0.3617, 50) = 0.64045

p-value > 0.25

Study 2: p-value = T.DIST.RT(1.5974, 21) = 0.06256

0.05 < p-value < 0.10