Question

Hi, can you help me answer this question please, thank you!

Answer 1

Given data:

Normally distributed and unknown population SD, variances are equal (assumed)

First Population

Sample Size = 12

Sample Mean = 75.4

Sample SD = 9.7

Second Population

Sample Size = 19

Sample Mean = 83.3

Sample SD = 17.8

Find: test statistic and p-value

Solution:

Since the sample size is small for both population and its standard deviation is unknown, we can use t-test. The formula for this is:

`t=\frac{\bar{x_1}-\bar{x_2}}{\sqrt[]{\frac{(n_1-1)s^2_{1^{}}+(n_2-1)s^{2_{}}_2}{n_1+n_2-2}((1)/(n_1)+(1)/(n_2))}};withdf=n_1+n_2-2`

Substitute the given data that we have above to the formula.

`\begin{gathered} t=\frac{75.4-83.3}{\sqrt[]{((12-1)(9.7)^2+(19-1)(17.8)^2)/(12+19-2)((1)/(12)+(1)/(19))}} \\ t=\frac{-7.9}{\sqrt[]{((11)(94.09)+(18)(316.84))/(29)((31)/(228))}} \\ t=\frac{-7.9}{\sqrt[]{(1034.99+5703.12)/(29)((31)/(228))}} \\ t=\frac{-7.9}{\sqrt[]{(6738.11)/(29)((31)/(228))}} \\ t=(-7.9)/(5.6206) \\ t\approx-1.406 \end{gathered}`

Hence, the test statistic is approximately -1.406.

The p-value for this test statistic is equal to 0.0852.