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Dual-energy X-ray absorptiometry (DXA) is a technique for measuring bone health. One of the most common measures is total body b

sesenic [268]

Answer:

$\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095$

-The sample is too small to make judgments about skewness or symmetry.

H0: $\mu_{1}=\mu_{2}$

H1: $\mu_{1} \neq \mu_{2}$

$t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013$

$p_v =2*P(t_{(14)}$

So the p value is a very high value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a we don't have a significant difference between the two means.

Step-by-step explanation:

First we need to find the difference defined as:

(Operator 1 minus Operator 2)

d1=1.326-1.323=0.003 d2=1.337-1.322=0.015

d3=1.079-1.073=0.006 d4=1.229-1.233=-0.004

d5=0.936-0.934=0.002 d6=1.009-1.019=-0.01

d7=1.179-1.184=-0.005 d8=1.289-1.304=-0.015

Now we can calculate the mean of differences given by:

$\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001$

And for the sample deviation we can use the following formula:

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095$

Describe the distribution of these differences using words. (which one is correct)

We can plot the distribution of the differences with the folowing code in R

differences<-c(0.003,0.015,0.006,-0.004,0.002,-0.01,-0.005,-0.015)

hist(differences)

And we got the image attached. And we can see that the distribution is right skewed but we don't have anough info to provide a conclusion with just 8 differnences.

-The sample is too small to make judgments about skewness or symmetry.

Use a significance test to examine the null hypothesis that the two operators have the same mean. Give the test statistic. (Round your answer to three decimal places.)

$\bar X_{1}=1.173$ represent the mean for the operator 1

$\bar X_{2}=1.174$ represent the mean for the operator 2

$s_{1}=0.1506$ represent the sample standard deviation for the operator 1

$s_{2}=0.1495$ represent the sample standard deviation for the operator 2

$n_{1}=8$ sample size for the operator 1

$n_{2}=8$ sample size for the operator 2

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

H0: $\mu_{1}=\mu_{2}$

H1: $\mu_{1} \neq \mu_{2}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We can replace in formula (1) like this:

$t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013$

Statistical decision

For this case we don't have a significance level provided $\alpha$ , but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=8+8-2=14$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{(14)}$

So the p value is a very high value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a we don't have a significant difference between the two means.

8 0

3 years ago

Ursula and Andre competed against each other in a swimming race. Ursula swam at a rate of 60 meters per minute and Andre swam at

Lubov Fominskaja [6]

Given that the two swimmers competed and Ursula's speed is 60 m/min while Andre's speed is 48 m/min. The distance that the Ursula will catch up with Andre will be:
distance=(relative speed)×(time)
relative speed=60-48=12 m/min
the two swimmers met at a distance of:
12×1
=12 meters

6 0

3 years ago

I WILL REPORT IF IT'S A LINK

Anit [1.1K]

Answer:

a) There will be approximately a water depth of 7.5 meters after 3h

b) Approximately 2.5 hours have passed when there is 8 m of water in the tank.

Step-by-step explanation:

a) Look for the y-value when the x-value is 3.

b) Look for the x-value when the y-value is 8.

*The y-axis is the depth (m), and the x-axis is the time (h).

7 0

2 years ago

You are testing the claim that the mean GPA of night students is different from the mea GPA of day students. You sample 20 night

svetlana [45]

Answer:

As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.

Step-by-step explanation:

Here n= 20

Sample mean GPA = x`= 2.84

Standard mean GPA = u= 2.55

Standard deviation = s= 0.45.

Level of Significance.= ∝ = 0.01

The hypothesis are formulated as

H0: u1=u2 i.e the GPA of night students is same as the mean GPA of day students

against the claim

Ha: u1≠u2

i.e the GPA of night students is different from the mea GPA of day students

For two tailed test the critical value is z ≥ z∝/2= ± 2.58

The test statistic

Z= x`-u/s/√n

z= 2.84-2.55/0.45/√20

z= 0.1441

As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.

8 0

3 years ago

What is the answer to this question ? 3a-4=5

Nataly_w [17]

Answer:

a=3

Step-by-step explanation:

a=3 because 5+4=9 and 9÷3=3

hope this helps!!

8 0

3 years ago

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