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3 years ago

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Mathematics

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Afina-wow [57]3 years ago

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The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who w

dimulka [17.4K]

Answer:

99% CI:

$-4.637\leq\mu_v-\mu_o\leq 3.737$

Step-by-step explanation:

We have to calculate a 99%CI for the difference of means for the vegan and the omnivore.

First, we have to estimate the standard deviation

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}$

The MSE can be calculated as

$MSE=\frac{(SSE_1+SSE_2)}{df} =\frac{(n_1s_1)^2+(n_2*s_2)^2}{n_1+n_2-2}\\\\MSE=\frac{(85*1.05)^2+(96*1.20)^2}{85+96-2}=\frac{7965.56+13272.04}{179} =118.65$

The weighted sample size nh can be calculated as

$n_h=\frac{2}{1/n_1+1/n_2}=\frac{2}{1/85+1/96}=\frac{2}{0.0222} =90.16$

The standard deviation then becomes

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}=\sqrt{\frac{2*118.65}{90.16}}=\sqrt{2.632} =1.623$

The z-value for a 99% confidence interval is z=2.58.

The difference between means is

$\Delta M=M_v-M_o=5.10-5.55=-0.45$

Then the confidence interval can be constructed as

$\Delta M-z*s_{Md}\leq\mu_v-\mu_o\leq \Delta M+z*s_{Md}\\\\ -0.45-2.58*1.623\leq\mu_v-\mu_o\leq -0.450+2.58*1.623\\\\-0.450-4.187\leq\mu_v-\mu_o\leq-0.450+4.187\\\\-4.637\leq\mu_v-\mu_o\leq 3.737$

8 0

3 years ago

Two increased by the product of a number and 9 is greater than or equal to 25.

Marianna [84]

Answer:

x≥2.56

Step-by-step explanation:

first we can write the inequality

2+9x≥25

we can now solve this with algebra

9x≥23

x≥2.56

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3 years ago

Unit 5 lesson 4.3 quiz answers

son4ous [18]

Answer:

There is nothing here to answer

Step-by-step explanation:

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3 years ago

Which table represents a direct variation function?

leonid [27]

3 i think it might be

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3 years ago

Read 2 more answers

Three consecutive even integers have a sum of 48 . find the integers.

Mandarinka [93]

x+(x+2)+(x+4)=48

3x+6=48

Minus 6 to both sides

3x=42

Divide 3 to both sides

x=14

Get the next two integers

x,x+2,x+4 or 14,16,18 = 48 is you Answer

4 0

3 years ago