Mathematics, 30.11.2019 02:31 shealwaysknows23

Pooled‐variances t‐test for mean comparison we have two normal populations with unknown means μ₁ and μ₂ and a common variance σ². two independent samples are randomly drawn from the populations with the following data: n₁ = 20 xത₁ = 8.91 s₁ = 2.1n₂ = 24 xത₂ = 8.23 s₂ = 2.5(a)[5] at the level of significance α = 0.01, test h₀: μ₁ = μ₂ versus h₁: μ₁ ≠ μ₂. sketch the test.

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With the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 1% of significance the mean of the group 1 is not significantly different from than the mean for the group 2.

Step-by-step explanation:

1) Notation and hypothesis

When we have two independentt 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:

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

This last one is an unbiased estimator of the common variance $\simga^2$

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2 \neq 0$

Our notation on this case :

n_1 =20 represent the sample size for group 1

n_2 =24 represent the sample size for group 2

$\bar X_1 =8.91$ represent the sample mean for the group 1

$\bar X_2 =8.23$ represent the sample mean for the group 2

s_1=2.1 represent the sample standard deviation for group 1

s_2=2.5 represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(20-1)(2.1)^2 +(24 -1)(2.5)^2}{20 +24 -2}=5.418$

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

S_p=2.328

2) Calculate the statistic

And now we can calculate the statistic:

$t=\frac{(8.91-8.23)-(0)}{2.328\sqrt{\frac{1}{20}}+\frac{1}{24}}=0.965$

Now we can calculate the degrees of freedom given by:

df=20+24-2=42

3) Calculate the p value

And now we can calculate the p value using the altenative hypothesis:

$p_v =2*P(t_{42}0.965) =0.340$

4) Conclusion

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 1% of significance the mean of the group 1 is not significantly different from than the mean for the group 2.

Answer from: Quest

step-by-step explanation:

Answer from: Quest

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