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3 years ago

What is the difference of the medians as a multiple of Megan's interquartile range?

2 answers:

6 0

From the figure associated to your question (which you did not attach), Amy's median is 40 and Megan's median is 35.

Thus, the difference in medians is given by 40 - 35 = 5.

From the figure, Megan's Q1 is 25 and her Q3 is 40, thus her interquatile range is given by Q3 - Q1 = 40 - 25 = 15.

Therefore, the difference of the medians as a multiple of Megan's interquartile range is given by

Difference in medians = 1/3 (Megan's IQR)

Fiesta28 [93]3 years ago

4 0

Answer: 1/3

Step-by-step explanation:

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2 years ago

Find the equation of the line that is parallel to x=5y-8 and pass through (6, -5) in slope intercept form

DochEvi [55]

Answer:

y=1/5x-31/5 (or y=0.2x-6.2 in decimals)

Step-by-step explanation:

x=5y-8 (original equation)

x+8=5y (Addition Property of Equality)

1/5x+8/5=y (Division Property of Equality, slope of original equation is 1/5)

y-y1=m(x-x1) (point-slope formula)

y-(-5))=1/5(x-(6)) (plug in the slope that was found earlier and the point given in the question)

y+5=1/5(x-6)

y+5=1/5x-6/5 (Distributive Property of Multiplication) Note: 5 = 25/5

y=1/5x-31/5 (Subtraction Property of Equality, and there's your answer)

y=0.2x-6.2 (This is the same answer, but written with decimals)

5 0

2 years ago

In a random sample of 75 American women age 18 to 30, 26 agreed with the statement that a woman should have the right to a legal

ddd [48]

Answer:

a) $z=\frac{0.347-0.328}{\sqrt{0.338(1-0.338)(\frac{1}{75}+\frac{1}{64})}}=0.236$

$p_v =2*P(Z>0.236)=0.813$

If we compare the p value and using any significance level for example $\alpha=0.01$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that we don't have significant differences between the two proportions.

b) We are confident at 99% that the difference between the two proportions is between $-0.188 \leq p_B -p_A \leq 0.226$

Step-by-step explanation:

Previous concepts and data given

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion of women age 18 to 30 agreed with the statement that a woman should have the right to a legal abortion for any reason

$\hat p_A =\frac{26}{75}=0.347$ represent the estimated proportion of women age 18 to 30 agreed with the statement that a woman should have the right to a legal abortion for any reason

$n_A=75$ is the sample size for A

$p_B$ represent the real population proportion for women age 58 to 70 agreed with the statement that a woman should have the right to a legal abortion for any reason

$\hat p_B =\frac{21}{64}=0.328$ represent the estimated proportion of women age 58 to 70 agreed with the statement that a woman should have the right to a legal abortion for any reason

$n_B=64$ is the sample size required for B

$z$ represent the critical value for the margin of error and for the statisitc

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Part a

We need to conduct a hypothesis in order to check if the proportion are equal, the system of hypothesis would be:

Null hypothesis: $p_{A} = p_{B}$

Alternative hypothesis: $p_{A} \neq p_{B}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{A}-p_{B}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{A}}+\frac{1}{n_{B}})}}$ (1)

Where $\hat p=\frac{X_{A}+X_{B}}{n_{A}+n_{B}}=\frac{26+21}{75+64}=0.338$

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.347-0.328}{\sqrt{0.338(1-0.338)(\frac{1}{75}+\frac{1}{64})}}=0.236$

Statistical decision

Since is a two sided test the p value would be:

$p_v =2*P(Z>0.236)=0.813$

Part b

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$(0.347-0.328) - 2.58 \sqrt{\frac{0.347(1-0.347)}{75} +\frac{0.328(1-0.328)}{64}}=-0.188$

$(0.347-0.328) + 2.58 \sqrt{\frac{0.347(1-0.347)}{75} +\frac{0.328(1-0.328)}{64}}=0.226$

And the 99% confidence interval for the difference of proportions would be given (-0.188;0.226).

We are confident at 99% that the difference between the two proportions is between $-0.188 \leq p_B -p_A \leq 0.226$

5 0

3 years ago

Choose all options that apply. Which of the following are equal to 20%? | a) .25 b) 1/5 Oc) 1/10 d) .20

zloy xaker [14]

Answer:

b. and d.

Step-by-step explanation:

$20 = \frac{20}{100} \\ 0.25 = \frac{25}{100} \\ \frac{1}{5} = \frac{20}{100} \\ \frac{1}{10} = \frac{10}{100} \\ 0.20 = \frac{20}{100}$

4 0

2 years ago

These are the round-trip distances that city buses traveled on different routes each month. Calculate the distance, in miles, th

prohojiy [21]

I think you have to use the distance formula

4 0

3 years ago