8.8 Small Sample Confidence Intervals(CI’s) for µ and µ1 - µ2 If Yi , i = 1, …,n, represents random sampling from a normal population where V(Yi ) = σ2 then the random variable T given by T = ( Y - µ) / (s/ √ n ) has a T distribution with (n - 1) degrees of freedom. T is called a pivotal quantity. For given α we want to find values ± tα/2 so that P( -tα/2 ≤ T ≤ tα/2 ) = 1 - α . The upper and lower limits are given by 1-α y ± tα/2 ( s / √ n ) -tα/2 0 tα/2 Example 8.11 p411. A manufacturer of gunpowder developed a new powder. Eight shells with the new powder were tested. The resulting muzzle velocitieds were : 3005 2995 2925 3005 2935 2937 2965 2905 Construct a 95% CI fot the average muzzle velocities of shells with the new powder. Assume muzzle velocities are normally distributed. Sol. The limits are given by y ± tα/2 ( s / √ n ). Here y = 2959, s = 39.1, n – 1 = 7, α/2 = .025, Therefore tα/2 = t.025 = 2.365. y ± tα/2 ( s / √ n ) = 2959 ± (2.365)(39.1 / √ 8 ) = 2926.3 or 2991.7. Thus the CI (2926.3, 2991.7) contains the average muzzel velocity with confidence coefficient .95 Comparing The Means of Two Normal Populations. When comparing the means of two populations for small samples (n < 30) we use the basic assumptions: 1. the samples are independent 2. the variances are equal ie. σ1 = σ2 = σ, but unknown 3. the sample sizes are n1 , and n2 . A. In the large sample case, use Z as the pivotal quantity where Z= where σ1 = σ2 = σ. (Y1 - Y2) – (µ1 - µ2) √ (σ12 /n1 ) + (σ22 /n2 ) B. In the small sample case, n< 30, use T as the pivotal quantity where T= (Y1 - Y2) – (µ1 - µ2) √ (sp2 /n1 ) + (sp2 /n2 ) where sp2 is the pooled estimator of σ2 and sp2 = ((n1 -1)s12 + (n2 -1)s22 )/(n1 + n2 -2). If n1 = n2 = n then sp2 = (s12 + s22 )/2. And the CI limits for µ1 - µ2 are: (Y1 - Y2) ± tα/2 sp √ (1/n 1 ) + (1/n 2 ) where tα/2 is determined from the t-distribution with (n 1 + n 2 - 2) df’s. Example: Seasonal ranges (in hectares) for alligators were monitored by biologists. Five alligators monitored in spring had ranges of 8.0, 12.1, 8.1, 18.2, 31.7. Four different alligators were monitored in Summer and had ranges of 102.0, 81.7, 54.7, 50.7. Use a CI with confidence 95% to estimate the difference in mean spring and summer ranges. What assumptions are made? Sol. Compute Y = 15.6, s = 9.90 and n = 5 Y = 72.3, s = 24.13 and n = 4 df’s = 5 + 4 – 2 = 7 1 - α = .95 α = .05 α/2 = .025 tα/2 = 2.365 sp2 = (4(9.90) + 3(24.13)) / (5 + 4 –2) or sp =3.99. Thus the limits are: (Y1 - Y2) ± tα/2 sp ± √ (1/n 1 ) + (1/n 2 ) = (1.56 - 72.3) ± (2.365)(3.99) √ (1/5) +(1/4) = -63.03 or -50.37. Thus a 95% CI is (-63.03, -50.37). The CI indicates the means are different. The assumptions are that the data are from normal populations and the sampling is random.. Exercises. p404. 8.68, 8.70, 8.73, 8.74, 8.76.