Question

Let μ1 and μ2 denote true average densities for two different types of brick. Assuming normality of the two density distributions, test H0: μ1 − μ2 = 0 versus Ha: μ1 − μ2 ≠ 0 using the following data: n1 = 8, x1 = 24.74,s1 = 0.161, n2 = 6, x2 = 20.92, and s2 = 0.250.
1. The degrees of freedom according to the satterthwaite approximation are:

2. Use α = 0.05. Round your test statistic to three decimal places and your P-value to four decimal places.) Use the satterthwaite degrees of freedom for your p-value.

P-value = ?

3. The point estimate and margin of error for the 95% confidence interval for μ1 − μ2 are: (Use the satterthwaite degrees of freedom for your critical value.)

Answer 1

Answer:

a) $df= \frac{(\frac{0.161^2}{8} +\frac{0.25^2}{6})^2}{\frac{1}{8 -1} (\frac{0.161^2}{8})^2 +\frac{1}{6 -1} (\frac{0.25^2}{6})^2} = 8.03 \ approx 8$

b) $t=\frac{24.74-20.92}{\sqrt{\frac{0.161^2}{8}+\frac{0.25^2}{6}}}=32.688$  

$p_v = 2*P(t_{8}>32.688)= 4.18x10^{-9}$

c) $ME= 2.306 \sqrt{\frac{0.161^2}{8}+\frac{0.25^2}{6}}= 0.269$

Step-by-step explanation:

Part a

For this case the satterthwaite approximation is given by:

$df= \frac{(\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2})^2}{\frac{1}{n_1 -1} (\frac{s^2_1}{n_1})^2 +\frac{1}{n_2 -1} (\frac{s^2_2}{n_2})^2}$

And if we replace we got:

$df= \frac{(\frac{0.161^2}{8} +\frac{0.25^2}{6})^2}{\frac{1}{8 -1} (\frac{0.161^2}{8})^2 +\frac{1}{6 -1} (\frac{0.25^2}{6})^2} = 8.03 \ approx 8$

Part b

Data given and notation  

$\bar X_{1}=24.74$ represent the mean for 1

$\bar X_{2}=20.92$ represent the mean for 2

$s_{1}=0.161$ represent the sample standard deviation for 1

$s_{2}=0.25$ represent the sample standard deviation for 2

$n_{1}=8$ sample size for 1

$n_{2}=6$ sample size for 2

t would represent the statistic (variable of interest)  

$p_v$ represent the p value  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for men it's higher than the mean for women, the system of hypothesis would be:  

H0:$\mu_{1} = \mu_{2}$  

H1:$\mu_{1} \neq \mu_{2}$  

If we analyze the size for the samples both are lower than 30, and we don't know the population deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:  

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)  

T-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

Calculate the statistic  

We have all in order to replace in formula (1) like this:  

$t=\frac{24.74-20.92}{\sqrt{\frac{0.161^2}{8}+\frac{0.25^2}{6}}}=32.688$  

P value

For this case since we have a bilateral test and since we know the degrees of freedom calculated form part a we can find the p value like this:

$p_v = 2*P(t_{8}>32.688)= 4.18x10^{-9}$

And since the $p_v < \alpha$ we have enough evidence to reject the null hypothesis at 5% of significance.

Part c

The margin of error for this case is given by:

$ME= t_{\alpha/2} \sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}$

We can find the critical value with the following excel formula:

"-=T.INV(0.025;8)"

$t_{\alpha/2}= 2.306$

And if we replace we got:

$ME= 2.306 \sqrt{\frac{0.161^2}{8}+\frac{0.25^2}{6}}= 0.269$