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2 years ago

Unit 4 homework 6 Gina Wilson

Mathematics

1 answer:

diamong [38]2 years ago

3 0

Answer:

Step-by-step explanation:

1) BD bisects ∠ABC Given

2) ∠BDA ≅ ∠BDC Given

3) BD ≅ BD Reflexive property

4)Δ BDA ≅ ΔBDC A S A congruent

2) 1.ST ≅ UV Given

2.∠WST ≅ ∠WVU Alternate interior angles, ST //UV and SV transversal

3. ∠WTS≅∠WUV Alternate interior angles, ST//UV and Tu transversal.

4. ΔWST ≅ΔWUV A S A

3) ∠EHF ≅ ∠GHF Given FH bisects ∠EHG

2.∠ FEH ≅∠FGH Given

FH ≅ FH Reflexive property

ΔFEH ≅ΔFGH A A S

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Null hypothesis: $\mu_1 \geq \mu_2$

Alternative hypothesis: $\mu_1$

$t=-1.329$

$p_v =P(t_{30}$

With the p value obtained and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the mena of the group 1 is significantly lower than the mean for the group 2.

Step-by-step explanation:

When we have two independnet 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: $\mu_1 \geq \mu_2$

Alternative hypothesis: $\mu_1$

Or equivalently:

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

Alternative hypothesis: $\mu_1 -\mu_2$

Our notation on this case :

$n_1 =14$ represent the sample size for group 1

$n_2 =18$ represent the sample size for group 2

$\bar X_1 =565$ represent the sample mean for the group 1

$\bar X_2 =578$ represent the sample mean for the group 2

$s_1=28.9$ represent the sample standard deviation for group 1

$s_2=26.3$ represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(14-1)(28.9)^2 +(18 -1)(26.3)^2}{14 +18 -2}=753.882$

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

$S_p=27.457$

And now we can calculate the statistic:

$t=\frac{(565 -578)-(0)}{27.457\sqrt{\frac{1}{14}}+\frac{1}{18}}=-1.329$

Now we can calculate the degrees of freedom given by:

$df=14+18-2=30$

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

$p_v =P(t_{30}$

So with the p value obtained and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the mena of the group 1 is significantly lower than the mean for the group 2.

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