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Karolina [17]
3 years ago
11

Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and st

andard deviation 1.

Mathematics
1 answer:
SashulF [63]3 years ago
7 0

Answer:

0.7233

Step-by-step explanation:

We want to find the area between the z-scores z=-0.95 and z=1.25.

We first find the area to the left of each z-score, and subtract the smaller area from the bigger one.

For the area to the left of z=-0.95, we read -0.9 under 5 from the standard normal distribution table.

This gives P(z<-0.95)=0.1711

Similarly the area to the left of z=1.25 is

P(z<1.25)=0.8944

Now the area between the two z-scores is

P(-0.25<z<1.25)=0.8944-0.1711=0.7233

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Answer:

With outlier (102)

t=\frac{126.2-125}{\frac{9.138}{\sqrt{10}}}=0.415      

p_v =P(t_{9}>0.415)=0.344    

If we compare the p value and a significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 125 feet at 5% of significance.    

Without outlier (102)

t=\frac{128.89-125}{\frac{3.551}{\sqrt{9}}}=3.285      

p_v =P(t_{8}>3.285)=0.0056  

And we conclude that we reject the null hypothesis since p_v. So the final conclusion would be not use the method since the value of 102 observed can be a potential outlier, removing this value we see that we reject the null hypothesis and we have a significant result that the true mean is higher than 125.

Step-by-step explanation:

Previous concepts  and data given  

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".  

Data: 129, 128, 130, 132, 135, 123, 102, 125, 128, 130

We can calculate the mean with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

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

\bar X=126.2 represent the sample mean    

s=9.138 represent the sample standard deviation  

n=10 represent the sample selected  

\alpha significance level    

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is significantly higher than 125, the system of hypothesis would be:    

Null hypothesis:\mu \leq 125    

Alternative hypothesis:\mu > 125    

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic  

We can replace in formula (1) the info given like this:    

t=\frac{126.2-125}{\frac{9.138}{\sqrt{10}}}=0.415      

P-value  

First we need to calculate the degrees of freedom given by:

df=n-1=10-1= 9

Then since is a right tailed sided test the p value would be:    

p_v =P(t_{9}>0.415)=0.344    

Conclusion    

If we compare the p value and a significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 125 feet at 5% of significance.    

Using all the data we see that we don;t have enough info to conclude that the true mean is higher than 125, but if we see careful 9 of the 10 values are over the limit of 125 feet. And if we repeat the procedure with the outlier of 102. We got this:

\bar X=128.89 represent the sample mean    

s=3.551 represent the sample standard deviation

t=\frac{128.89-125}{\frac{3.551}{\sqrt{9}}}=3.285      

First we need to calculate the degrees of freedom given by:

df=n-1=9-1= 8

Then since is a right tailed sided test the p value would be:    

p_v =P(t_{8}>3.285)=0.0056  

And we conclude that we reject the null hypothesis since p_v. So the final conclusion would be not use the method since the value of 102 observed can be a potential outlier removing this value we see that we reject the null hypothesis and we have a significant result that the true mean is higher than 125.

8 0
3 years ago
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