To which place value is the number rounded?

The output from a statistical computer program indicates that the mean and standard deviation of a data set consisting of 200 me

lesya [120]

Answer:

The limit that 97.5% of the data points will be above is $912.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 1500, \sigma = 300$

Find the limit that 97.5% of the data points will be above.

This is the value of X when Z has a pvalue of 1-0.975 = 0.025. So it is X when Z = -1.96.

So

$Z = \frac{X - \mu}{\sigma}$

$-1.96 = \frac{X - 1500}{300}$

$X - 1500 = -1.96*300$

$X = 912$

The limit that 97.5% of the data points will be above is $912.

6 0

3 years ago

PLEASEE HELP!!<br><br> Write and equation for this graph in slope- intercept form.

ella [17]

-2/1. You find two point that go on the line and if it’s going down it’s negative

8 0

2 years ago

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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

Una cámara fotográfica tiene un precio 5200 sin iva ¿cuanto se pagará de iva 16%

Free_Kalibri [48]

Answer:

el tiempo es de 46% caracteres fotográficos ya que si el precio es de 5200 la mitad es 3890

6 0

3 years ago

Tiana can clear a football field of debris in 3 hours. Jacob can clear the same field in 2 hours. If they decide to clear the fi

Lapatulllka [165]

Let us determine each of their rates of work:

Since Tiana can clear the field in 3 hours, meaning she can do 1/3 of the work per hour.

In the same way, Jacob can do 1/2 of the work per hour.

Now, what would be the rate of work if they were working together? Let's look at it like this:

Hours to complete the job:
Tiana = 3
Jacob = 2
Together = t

Work done per hour:
Tiana = 1/3
Jacob = 1/2
Together = 1/t

If you add their labor together:

1/3 + 1/2 = 1/t
5/6 = 1/t
t = 6/2 = 1.2

Together, they can clear the field in 1.2 hours.

4 0

3 years ago

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