Refer to right triangle ABC with C = 90°. If a = 6 and c = 14, find b.

The accounting department analyzes the variance of the weekly unit costs reported by two production departments. A sample of 16

dusya [7]

Answer:

$F=\frac{s^2_2}{s^2_1}=\frac{5.5}{2.5}=2.2$

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have $n_2 -1 =16-1=15$ and for the denominator we have $n_1 -1 =16-1=15$ and the F statistic have 15 degrees of freedom for the numerator and 15 for the denominator. And the P value is given by:

$p_v =2*P(F_{15,15}>2.2)=0.138$

For this case the p value is highert than the significance level so we haev enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviations are not significantly different

Step-by-step explanation:

Information given

$n_1 = 16$ represent the sampe size 1

$n_2 =16$ represent the sample 2

$s^2_1 = 2.5$ represent the sample deviation for 1

$s^2_2 = 5.5$ represent the sample variance for 2

$\alpha=0.10$ represent the significance level provided

The statistic is given by:

$F=\frac{s^2_2}{s^2_1}$

Hypothesis to test

We want to test if the variations in terms of the variance are equal, so the system of hypothesis are:

H0: $\sigma^2_1 = \sigma^2_2$

H1: $\sigma^2_1 \neq \sigma^2_2$

The statistic is given by:

$F=\frac{s^2_2}{s^2_1}=\frac{5.5}{2.5}=2.2$

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have $n_2 -1 =16-1=15$ and for the denominator we have $n_1 -1 =16-1=15$ and the F statistic have 15 degrees of freedom for the numerator and 15 for the denominator. And the P value is given by:

$p_v =2*P(F_{15,15}>2.2)=0.138$

For this case the p value is highert than the significance level so we haev enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviations are not significantly different

5 0

3 years ago

The price of 4 dresses is $486 What is the rate of each dress?

Mariulka [41]

D. is the correct answer your just divide 486 by 4 you get 121.50

4 0

2 years ago

Historically, the proportion of people who trade in their old car to a car dealer when purchasing a new car is 48%. Over the pre

choli [55]

Answer:

$z=\frac{0.4 -0.48}{\sqrt{\frac{0.48(1-0.48)}{115}}}=-1.717$

$p_v =P(z$

So the p value obtained was a very low value 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 5% of significance the proportion of people that have traded in their old car is lower than 0.48 or 48%.

Step-by-step explanation:

Data given and notation

n=115 represent the random sample taken

X=46 represent the number of people that have traded in their old car.

$\hat p=\frac{46}{115}=0.4$ estimated proportion of people that have traded in their old car

$p_o=0.48$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.9

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.48.:

Null hypothesis: $p\geq 0.48$

Alternative hypothesis: $p < 0.48$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.4 -0.48}{\sqrt{\frac{0.48(1-0.48)}{115}}}=-1.717$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.1$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(z$

So the p value obtained was a very low value 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 5% of significance the proportion of people that have traded in their old car is lower than 0.48 or 48%.

4 0

3 years ago

Can somebody help me

Sergeeva-Olga [200]

$cis(x)=\cos x+i\sin x\\\\cis15^o=\cos15^o+i\sin15^o\\\\\cos15^o=\dfrac{\sqrt6+\sqrt2}{4};\ \sin15^o=\dfrac{\sqrt6-\sqrt2}{4}\\\\\text{therefore}\\\\7cis15^o=7\left(\cos15^o+i\sin15^o\right)=7\cos15^o+7i\sin15^o\\\\=7\left(\dfrac{\sqrt6+\sqrt2}{4}\right)+7\left(\dfrac{\sqrt6-\sqrt2}{4}\right)i$

6 0

3 years ago

The graph shown compares the rent Andrew and Dave pay for renting bikes from different stores

Dmitry_Shevchenko [17]

Answer:I don’t know either

Step-by-step explanation:

5 0

3 years ago

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