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san4es73 [151]
3 years ago
14

Suppose a scatter plot contains the points (12, 28), (15, 33), (21, 44), (29, 62), and (33, 68). Which of these points would be

considered an outlier if it were also on the scatter plot? Select three that apply.
Mathematics
1 answer:
yawa3891 [41]3 years ago
6 0

Answer:

Unsure, take a look at the visualisation:

desmos(dotcom)/calculator/fjze0od81o

Replace the (dotcom) with ".com"    (   links are not allowed :(   )

Explanation:

None of the points look as though they are an outlier from the line of linear regression, although (29,62) looks furthest from the line so could be a possible answer although I would probably say there is no outlier.

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What number is 6 more than 7?
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A lathe is set to cut bars of steel into lengths of 6 cm. The lathe is considered to be in perfect adjustment if the average len
Gnom [1K]

Answer:

t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723    

E. -0.723

df=n-1=93-1=92  

p_v =2*P(t_{(92)}  

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

Step-by-step explanation:

Information provided

\bar X=5.97 represent the sample mean for the length

s=0.4 represent the sample standard deviation

n=93 sample size  

\mu_o =6 represent the value that we want to test

\alpha=0.05 represent the significance level

t would represent the statistic  

p_v represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the lathe is in perfect adjustment (6cm), then the system of hypothesis would be:  

Null hypothesis:\mu = 6  

Alternative hypothesis:\mu \neq 6  

since we don't know the population deviation the statistic is:

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

Replacing in formula (1) we got:

t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723    

E. -0.723

P value

The degrees of freedom are given by:

df=n-1=93-1=92  

Since is a two tailed test the p value would be:  

p_v =2*P(t_{(92)}  

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

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2 years ago
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