Molly rewrote the equation as shown:

Sketch the area under the standard normal curve over the indicated interval and find the specified area. (Round your answer to f

Angelina_Jolie [31]

Answer:

$P(z>1.51) =1-P(Z$

And i fwe look into the normalstandar dtable we can find the desired probability:

$P(z>1.51) =1-P(Z$

And then the The area to the right of z = 1.51 is 0.0655

Step-by-step explanation:

For this case we want to find the following probability:

$P(z>1.51)$

And for this case we can use the normal standard distribution and the complement rule and we have this:

$P(z>1.51) =1-P(Z$

And i fwe look into the normalstandar dtable we can find the desired probability:

$P(z>1.51) =1-P(Z$

And then the The area to the right of z = 1.51 is 0.0655

7 0

3 years ago

How do I solve 3/10+2.125 <br> Converted into fractions

ahrayia [7]

Answer:

97/40

Step-by-step explanation:

so 2.125= 17/8

3/10+17/8 need to find common denominator so

times 8/8 to 3/10

times 10/10 to 17/8

24/80+170/80= 194/80 (common denominator so can add)

divide 194/80 because want reduce form

97/40 is the answer

(When i look at it now i can just time 4/4 and 5/5 but its ok this still gets the answer)

3 0

3 years ago

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.