Math / Data & Statistics

QuestionFind the critical values, $t_{0}$ , to test the claim that $\mu_{1}=\mu_{2}$ . Two samples are randomly Submit test $\sigma_{1}^{2} \neq \sigma_{2}^{2}$ . Use $\alpha=0.05$ . $n_{1}=25, n_{2}=30, \bar{x}_{1}=16, \bar{x}_{2}=14, s_{1}=1.5, s_{2}=1.9$ A. $\pm 2.064$ B. $\pm 2.797$ $\pm 1.711$ $\pm 2.492$

Studdy Solution

STEP 1

1. We are conducting a two-sample t-test to compare the means of two independent samples.
2. The population variances are unequal ( $\sigma_{1}^{2} \neq \sigma_{2}^{2}$ ), so we will use the Welch's t-test.
3. The significance level is $\alpha = 0.05$ .
4. The sample sizes are $n_{1} = 25$ and $n_{2} = 30$ .
5. The sample means are $\bar{x}_{1} = 16$ and $\bar{x}_{2} = 14$ .
6. The sample standard deviations are $s_{1} = 1.5$ and $s_{2} = 1.9$ .

STEP 2

1. Calculate the degrees of freedom using the Welch-Satterthwaite equation.
2. Find the critical t-values for the calculated degrees of freedom and the given significance level.

STEP 3

Calculate the degrees of freedom using the Welch-Satterthwaite equation:
$df = \frac{\left( \frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}} \right)^{2}}{\frac{\left( \frac{s_{1}^{2}}{n_{1}} \right)^{2}}{n_{1} - 1} + \frac{\left( \frac{s_{2}^{2}}{n_{2}} \right)^{2}}{n_{2} - 1}}$
Substitute the given values:
$df = \frac{\left( \frac{1.5^{2}}{25} + \frac{1.9^{2}}{30} \right)^{2}}{\frac{\left( \frac{1.5^{2}}{25} \right)^{2}}{24} + \frac{\left( \frac{1.9^{2}}{30} \right)^{2}}{29}}$
Calculate:
$df \approx 50.08$

STEP 4

Find the critical t-values for $df \approx 50$ and $\alpha = 0.05$ (two-tailed test):
Using a t-distribution table or calculator, find the critical value for $df = 50$ and $\alpha/2 = 0.025$ :
$t_{0} \approx \pm 2.009$
The critical values are:
$\boxed{\pm 2.009}$

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