Question

Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The first sample has a mean of 43.5 and standard deviation of 4.1 while the second sample has a mean of 40.1 and standard deviation of 3.2. A researcher would like to test if there is a difference between the population means at the 0.05 significance level. What can the researcher conclude?

Answer 1

Answer:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that  

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

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

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

We have the following data given:

$n_1 =20$ represent the sample size for group 1

$n_2 =20$ represent the sample size for group 2

$\bar X_1 =43.5$ represent the sample mean for the group 1

$\bar X_2 =40.1$ represent the sample mean for the group 2

$s_1=4.1$ represent the sample standard deviation for group 1

$s_2=3.2$ represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525$

And the deviation would be just the square root of the variance:

$S_p=3.678$

The statistic is givne by:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different