Mathematics, 07.06.2020 00:02 lisaaprice14

A research company desires to know the mean consumption of meat per week among people over age 29. A sample of 2092 people over age 29 was drawn and the mean meet consumption was 2.9 pounds. Assume that the standard deviation is known to be 1.4 pounds. Construct a 95% confidence interval for the mean consumption of meat among people over age 29. Round your answer to one decimal place.

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Answer from: lilkit

The 95% confidence interval for the mean consumption of meat among people over age 29 is between 2.8 pounds and 3 pounds.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of 1-0.025 = 0.975 , so z = 1.96

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{1.4}{\sqrt{2092}} = 0.1$

The lower end of the interval is the sample mean subtracted by M. So it is 2.9 - 0.1 = 2.8 pounds

The upper end of the interval is the sample mean added to M. So it is 2.9 + 0.1 = 3 pounds.

The 95% confidence interval for the mean consumption of meat among people over age 29 is between 2.8 pounds and 3 pounds.

Answer from: Quest

see below

step-by-step explanation:

7. there are 10 data points in the set. the middle point is between 5 and 6. the middle point in the top half is 8 and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term point 8 = 100 point 3 = 88

iqr = 100-88 = 12

final point 8 = 93 point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6. b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical . there is an outlier for the males that skews the data

5. the iqr is a better measure of spread for the movies and basketball games. since the basketball game has an outlier, we do not want to use standard deviation for measurement.

3. the exam median is much higher than the class median. the median line is farther to the right on the exams

2. there is no way to tell what the means are from the box and whisker plots.

Answer from: Quest

yes

step-by-step explanation:

so of course you want to start by plugging the 5 in which goes to 7(5)+3(5)=50 because 7*5=35 and 3*5=15 and then add them together, then plug in the 5 to the other equation as well 9(5)+5=50 because 9*5= 45 and then add 5 and so yes they are equivalent.