Question

Inspired by zoom, you want to determine whether there are significant differences in the satisfaction of couples that had their first date in person vs. couples who had their first date virtually. Based on the following satisfaction data (larger values indicate greater satisfaction), conduct an independent samples t-test using (alpha=.01) and an pooled variance. Is there a significant difference?
Virtual Date In person date
X= 23.76 X= 34.54
s2 = 123.43 s2 = 105.65
n = 5 n = 20
A) Write the null and alternative hypotheses in symbols
B) What is your critical value?
C) What's your pooled variance?
D) What's your standard error?
E) What's your obtained t?

Answer 1

Answer:

A) The null hypothesis is H₀; $\overline x_1$ = $\overline x_2$

The alternative hypothesis is Hₐ; $\overline x_1$ ≠ $\overline x_2$

B) The critical value is 4.604

C) The pooled variance is 11,870.2777

D) The standard error is-0.2321

E) The obtained is -0.2040  

Step-by-step explanation:

The given values are;

$\overline x_1$ = 23.76, $\overline x_2$ = 34.54

s₁ = 123.43, s₂ = 105.65

n₁ = 5, n₂ = 20

A) The null hypothesis, H₀; $\overline x_1$ = $\overline x_2$

The alternative hypothesis, Hₐ; $\overline x_1$ ≠ $\overline x_2$

B) The degrees of freedom, df = 4 - 1 = 3

The critical t-value is given at α = 0.01 is given on the t-table for two tailed test as t = 4.604

C) The pooled variance is given by the following formula;

$S_P^2 = \dfrac{(n_1 - 1) \cdot s_1^2 + (n_2-1) \cdot s_2^2}{n_1 + n_2 - 2 }$

Plugging in the values gives;

$S_P^2 = \dfrac{(5 - 1) \cdot 123.43^2 + (20-1) \cdot 105.65^2}{5+ 20 - 2 } = 11,870.2777$

The pooled variance, $s_p^2$ = 11,870.2777

D) The standard error, SE, is given as follows;

$t=\dfrac{(\bar x_{1}-\bar x_{2})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}$

Therefore, we get;

$t=\dfrac{(23.76-35.34)}{\sqrt{\dfrac{123.43^{2} }{5}-\dfrac{105.65^{2}}{20}}} = -0.2321$

The standard error, SE ≈ -0.2321

E) The obtained 't' is given as follows;

$\sigma _{\overline x_1 - \overline x_2} = \sqrt{\dfrac{s_1^2}{n_1-1} +\dfrac{s_2^2}{n_2-1} }$

Therefore, we have;

$\sigma _{\overline x_1 - \overline x_2} = \sqrt{\dfrac{123.43^2}{5-1} +\dfrac{105.65^2}{20-1} } \approx 56.7564$

The obtained t is given as follows;

$t(Obtained)=\dfrac{(\bar x_{1}-\bar x_{2})}{\sigma _{\overline x_1 - \overline x_2} }$

Therefore, we get;

$t(Obtained)=\dfrac{(23.76-34.54)}{56.7564} \approx -0.2040$

The obtained t = -0.2040