Question

With In C. nt the variances are unchanged. first art b, the two samples are bigger than i sample ând s 76 for s 68 for the size re affect the size of the standard error for the n Part a, How does sample sample mean difference? ach of the following, calculate the pooled vari- 22F For each of the foll mean difference a 217, and the second sample has n 8 scores h. Now the sample variances are increased so that the eand the estimated standard error for the sample The first sample has and a variance of s2 27 n 4 scores and a variance of first sample has n 4 scores and a variance of 2 68, and the second sample has n 8 scores and a variance of s2 108. c. Comparing your answers for Parts a and b, how does increased variance influence the size of the estimated standard error? 23. It appears that there is some truth to the old adage "That which doesn't kill us makes us stronger." Seery Holman, and Silver (2010) found that individuals with some history of adversity report better mental health and higher well-being compared to people with litle no history of adversity. In an attempt to examine non, a researcher surveys a group of col nm

Answer 1

22)

The pooled variance is obtained using the formula,

$s_{p}^{2}={\frac {(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}$

$\Rightarrow \text{Where, }n_1=8,\ s_1^2=17,\ n_2=8\ \text{and } s_2^2=27$

$s_{p}^{2}={\frac {(4-1)17^{2}+(8-1)27^{2}}{4+8-2}}$

$\text{Pooled Variance, }s_{p}^{2}=24$

The standard error for difference in mean is obtained using the formula,

$SE_{\mu_1-\mu_2}=\sqrt{\frac{s_p^2}{n_1}+\frac{s_p^2}{n_2}}$

$SE_{\mu_1-\mu_2}=\sqrt{\frac{24}{4}+\frac{24}{8}}$

$\text{Standard Error, }SE_{\mu_1-\mu_2}=3$

b)

The pooled variance is obtained using the formula,

$s_{p}^{2}={\frac {(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}$

$\Rightarrow \text{Where, }n_1=8,\ s_1^2=68,\ n_2=8\ \text{and } s_2^2=108$

$s_{p}^{2}={\frac {(4-1)68^{2}+(8-1)108^{2}}{4+8-2}}$

$\text{Pooled Variance, }s_{p}^{2}=96$

The standard error for difference in mean is obtained using the formula,

$SE_{\mu_1-\mu_2}=\sqrt{\frac{s_p^2}{n_1}+\frac{s_p^2}{n_2}}$

$SE_{\mu_1-\mu_2}=\sqrt{\frac{96}{4}+\frac{96}{8}}$

$\text{Standard Error, }SE_{\mu_1-\mu_2}=6$

c)

The standard for part a and b are,

$\Rightarrow \text{For } s_1^2=17\ \text{and}\ s_2^2=27,\ \text{Standard Error, }SE_{\mu_1-\mu_2}=3$

$\Rightarrow \text{For } s_1^2=68\ \text{and}\ s_2^2=108,\ \text{Standard Error, }SE_{\mu_1-\mu_2}=6$

$\text{We can see that, }$

$s_1^2=17\times 4=68\ \text{and}\ s_2^2=27 \times 4= 108,\ \text{Standard Error, }SE_{\mu_1-\mu_2}=3\times 2=6$

$\mathbf{ \text{If each standard deviations multiplied by 4, standard error got doubled}}$

24)

Question: If other are held constant....

Solution:

The t statistic for the difference of means is obtained by using the formula,

${\displaystyle t={\frac {{\mu}_{1}-{\mu}_{2}}{SE_{\mu_1-\mu_2 }}}}$

a)

Here we can see that,

$t\ \text{is proportional to difference of means,} \ {\mu}_{1}-{\mu}_{2}$

$\text{If both the score increase by same amount,}$

$\Rightarrow \mu_1-\mu_2 \text{ will be unaffected hence t statistic will be unaffected}$

b)

$t\ \text{is inveresly proportional to Standard error,} \ \frac{1}{SE_{\mu_1-\mu_2}}$

$\text{If both the variance increase by same amount,}$

$\Rightarrow SE_{\mu_1-\mu_2} \text{ will decrease by square root of that amount}$

$\Rightarrow \text{ hence t statistic will increased by square root of that amount}$