Question

Listed below are the summary statistics from samples of strontium-90 in baby teeth obtained from twelve Pennsylvania residents and another twelve from New York residents born after 1979. PA NY Mean 147.583 136.417 Stdev 10.6383 5.21289 If we wished to test the claim that the mean amount of strontium-90 from Pennsylvania residents is greater than the mean of New York residents, and can assume that the variances are equal, what is the Test Statistic for this test

Answer 1

Answer:

9.72

Step-by-step explanation:

s1 = 10.6383 ; s2 = 5.21289

x1 = 147.583 ; x2 = 136.417

n1 = 12 ; n2 = 12

df1 = n1 - 1 = 12 - 1 = 11

df2 = n2 - 1 = 12 - 1 = 11

The test statistic :

(x1 - x2) / sqrt[(sp²/n1 + sp²/n2)]

Pooled variance = Sp² = (df1*s1² + df2*s2²) ÷ (n1 + n2 - 2)

Sp² = ((11*10.6383) + (11*5.21289)) / 22 = 7.926

Test statistic, T* :

(147.583 - 136.417) / √(7.926 * (1/12 + 1/12))

11.166 / √(7.926 * (1/6)

11.166 / √1.321

11.166 / 1.1493476

T* = 9.7150766

Test statistic = 9.72