this problem is from the wackerly, but i have a particular question, in this problem we have to estimate the variance with S p 2 or can we use the information standard deviations that they gave us.Here is the problem:Two new drugs were given to patients with hypertension. The ﬁrst drug lowered the blood pressure of 16 patients an average of 11 points, with a standard deviation of 6 points. The second drug lowered the blood pressure of 20 other patients an average of 12 points, with a standard deviation of 8 points. Determine a 95% conﬁdence interval for the difference in the mean reductions in blood pressure, assuming that the measurements are normally distributed with equal variances.

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this problem is from the wackerly, but i have a particular question, in this problem we have to estimate the variance with       S    p    2   or can we use the information standard deviations that they gave us.Here is the problem:Two new drugs were given to patients with hypertension. The ﬁrst drug lowered the blood pressure of 16 patients an average of 11 points, with a standard deviation of 6 points. The second drug lowered the blood pressure of 20 other patients an average of 12 points, with a standard deviation of 8 points. Determine a 95% conﬁdence interval for the difference in the mean reductions in blood pressure, assuming that the measurements are normally distributed with equal variances.

Valeria Wolfe · Accepted Answer

Assuming the given sample standard deviations       S    1    ,      S    2   have respective denominators 16−1 and 20−1, the pooled sample variance is                              S          p          2                                    =                              (            16            −            1            )                          S              1              2                        +            (            20            −            1            )                          S              2              2                                            (            16            −            1            )            +            (            20            −            1            )                          =                              (            15            ×                          6              2                        )            +            (            19            ×                          8              2                        )                                (            16            −            1            )            +            (            20            −            1            )                                                            =                              540            +            1216                    34                =                  1756          34                ≈                  7.186589          2                .            Recall that   (  (  16  −  1  )  +  (  20  −  1  )  )                    S        p        2                    σ        2              ∼      χ          (      16      −      1      )      +      (      20      −      1      )        2    .You need to use       S    p   with a t-distribution with   (  16  −  1  )  +  (  20  −  1  )  =  34 degrees of freedom.

this problem is from the wackerly, but i have a particular question, in this problem we have to esti

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