Samples are taken from two different types of honey and the viscosity is measured.Honey A:Mean: 114.44S.D : 0.62Sample Size: 4Honey B:Mean: 114.93S.D: 0.94Sample Size: 6Assuming normal distribution, test at 5% significance level whether there is a difference in the viscosity of the two types of honey?Here's what I did:I took my null hypothesis as μB - μA = 0 and alternative hypothesis as μB - μA ≠ 0Then I did my calculations which were as following:Test Statistic = (B -A ) - ( μB - μA) / sqrt {(variance B / sample size B) + (variance A / sample size A)}This gave me test statistic as = 0.49/0.49332 that is equal to 0.993However the test statistic in the book solution is given as 0.91. What am I doing wrong?

Question

Samples are taken from two different types of honey and the viscosity is measured.Honey A:Mean: 114.44S.D : 0.62Sample Size: 4Honey B:Mean: 114.93S.D: 0.94Sample Size: 6Assuming normal distribution, test at 5% significance level whether there is a difference in the viscosity of the two types of honey?Here&#039;s what I did:I took my null hypothesis as   μB -   μA = 0 and alternative hypothesis as   μB -   μA   ≠ 0Then I did my calculations which were as following:Test Statistic = (B -A ) - (  μB -   μA) / sqrt {(variance B / sample size B) + (variance A / sample size A)}This gave me test statistic as = 0.49/0.49332 that is equal to 0.993However the test statistic in the book solution is given as 0.91. What am I doing wrong?

candydulce168nlid · Accepted Answer

The book is using a pooled standard error, one of two possible methods. The central limit theorem says that                     x        ¯              1    −                    x        ¯              2   is approximately normal with mean       μ    1    −      μ    2   and standard deviation                     σ        1        2                    n        1              +                  σ        2        2                    n        2            , but when the population sizes are small and it can be reasonably assumed the populations have the same variance, then a pooled standard error is used with       s    p    =                    (                  n          1                −        1        )                  s          1          2                +        (                  n          2                −        1        )                  s          2          2                                      n          1                +                  n          2                −        2              =                              .62          3                ∗        3        +                  .94          2                ∗        5            8       giving a standard error of                     x        ¯              1    −                    x        ¯              2   of       s    p              1              n        1              +          1              n        2            . This is to combine more data into the calculation of the standard error of each population mean, which are assumed to be equal for these purposes.So the t-distribution with       n    1    +      n    2    −  2  =  8 degrees of freedom has test statistic      t    ∗    =            (                                    x            ¯                          1            −                                    x            ¯                          2            )      −      (              μ        1            −              μ        2            )                      s        p                              1          4                +                  1          6                      =  0.9096458

Samples are taken from two different types of honey and the viscosity is measured. Honey A:

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