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spin [16.1K]
2 years ago
6

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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Given the following null and alternative hypotheses H0: μ1 ≥ μ2 HA: μ1 < μ2 Together with the following sample information (s
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Answer:

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.  

7 0
3 years ago
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