Mathematics, 26.02.2020 04:32 jadenpmoore2008

An article presents results of a survey of adults with diabetes. The average body mass index (BMI) in a sample of 1551 men was 30.4, with a standard deviation of 0.6. The average BMI in a sample of 1920 women was 31.1 with a standard deviation of 0.2. Let μX represent the population mean BMI in women with diabetes and let μY represent the population mean BMI in men with diabetes. Find a 99% confidence interval for the difference μX−μY. Round the answers to three decimal places.

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Answer from: shelly74

$(31.1 -30.4) - 2.58 \sqrt{\frac{0.2^2}{1920}+\frac{0.6^2}{1551}}= 0.659$

$(31.1 -30.4) + 2.58 \sqrt{\frac{0.2^2}{1920}+\frac{0.6^2}{1551}}= 0.741$

So then the 99% confidence interval for $\mu_X -\mu_Y$ is givenby (0.659; 0.741)

Step-by-step explanation:

Previous concepts

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".

We have the following data given:

$\bar X = 31.1$ represent the sample mean for women

n_x = 1920 the sample size of women

s_x = 0.2 standard deviation for women

$\bar Y = 30.4$ represent the sample mean for men

n_y = 1551 the sample size of men

s_y = 0.6 standard deviation for men

Solution to the problem

The confidence interval is given by the following formula:

$(\bar X -\bar Y) + t_{\alpha/2} \sqrt{\frac{s^2_x}{n_x}+\frac{s^2_y}{n_y}}$ (1)

Since the sample sizes are large enough we can use the normal distribution as approximation of the t distribution. 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 formula (1) we got:

$(31.1 -30.4) - 2.58 \sqrt{\frac{0.2^2}{1920}+\frac{0.6^2}{1551}}= 0.659$

$(31.1 -30.4) + 2.58 \sqrt{\frac{0.2^2}{1920}+\frac{0.6^2}{1551}}= 0.741$

So then the 99% confidence interval for $\mu_X -\mu_Y$ is givenby (0.659; 0.741)

Answer from: Quest

$15 each

step-by-step explanation:

$45/3 children= $15 each

Answer from: Quest

when rotating a figure 180 degrees, imagine that you are able to take the figure and turn it clockwise or counterclockwise around a center point. to rotate a figure 180 degrees, you will need to apply the rule (x, y) → (-x, -y). start by using a coordinate grid with coordinates for each vertex of the figure. the center point of the coordinate grid is located at (0, 0), which is what you will rotate the figure around. write down the original coordinates of the shape you are going to rotate. then, apply the rule. for example, coordinate (1, 2) becomes (-1, -2). another example would be, coordinate (-4, -2) becoming (4, 2). after you change each original coordinate to the rotated coordinate, you will draw your new figure.

step-by-step explanation:

Identify the transformation that maps the figure with center (7, 1) onto itself. a) rotate 180° cloc

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