Question

The distributions with large n don't need to be normal, but these are.

In this problem you do not know the population standard deviation.

The mean for the first set ¯xx¯ ₁ = 12.346 with a standard deviation of s₁ = 2.215 There sample size was 14.

The mean for the first set ¯xx¯  2  = 12.181 with a standard deviation of s₂ = 1.888 There sample size was 21.

Find the degrees of freedom.   (round to 2 places) The formula is the messy one on the MOM chapter 10 instruction page.

Use a two tail test. Find the left tail critical value, t-star for alpha = 0.01. Use the truncated version of the degrees of freedom.   (Remember the right tail is just the positive version. Use the inverse of t on excel. You will get a negative number.)

Find the test statistic t=   Round to 4 places.

Answer 1

Answer:

Given Data

= 12.346

11

s₁ = 2.215

= 14

11

. = 12.181

22

s₂ = 1.888

=21.

2

We assume that unequal variances.

1) The degrees of freedom

  $df=\frac{\left [ \frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}} \right ]^{2}}{\frac{\left ( \frac{s_{1}^{2} }{n_{1}}\right )^{2}}{n_{1}-1}+\frac{\left ( \frac{s_{2}^{2} }{n_{2}}\right )^{2}}{n_{2}-1}}$

$=\frac{\frac{(2.25)^{2}}{14}+\frac{(1.888)^{2}}{21}}{\frac{\left ( \frac{5.0625}{14} \right )^{2}}{14-1}+\frac{\left ( \frac{(3.5645)}{21} \right )^{2}}{21-1}}$

$= \frac{0.36161+0.1697}{0.0101+0.00144}$

$= \frac{0.53131}{0.01154}$

= 46.0407279

  $df\cong 46$

2)  $\alpha = 0.01$

It is two tailed test .

$\alpha /2= 0.005$

$t_{\alpha/2,df}=t_{0.005,46}$

= $\pm 2.9487$

The Right critical value = + 2.9487

3) The test Statistic:

$t=\frac{\bar{X}_{1}-\bar{X_{2}}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}$

$=\frac{12.346-12.181}{\sqrt{\frac{4.906225}{14}+\frac{3.564544}{21}}}$

  $=\frac{0.165}{\sqrt{0.35044+0.16974}}$

$=\frac{0.165}{0.72124}$

= 0.22877

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