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Cerrena [4.2K]
4 years ago
9

50 POINTS!!!! Make a box and whisker plot of the data See photo

Mathematics
1 answer:
lara [203]4 years ago
5 0

Let's try it just using test taking skills first, without really knowing anything about what these plots are.

All have a minimum of 13 and a max of 29 so those don't help.

Three have a left box edge of 14.5, one 12.8 or so.  The correct answer is probably 14.5.

Three have a median of 19.5, one of 20.5, so 19.5 is probably right.

Similarly 23.5 is the most common right box edge.  

So we're looking for the plot with 13/14.5/19.5/23.5/29, which is the first choice.

----

Instead of Einsteining the test, lets review the material.  A box and whiskers plot shows the min and max (whisker ends), the 25th and 75th percentiles (box ends) and the 50th percentile, the median, which is the dot in the middle.

Let's start by sorting the 8 data points:

11 13 16 18 21 23 24 29

The minimum is 11 and the maximum is 29.  All the charts agree on this.

The median is the middle, which is the average of 18 and 21, so

median = 19.5

That rules out the one selected, it can't be the third one.

The 25th percentile basically the median of the first four numbers, so the average of 13 and 16,

25th percentile = 14.5

The 75th percentile is similarly the average of 23 and 24

75th percentile 23.5

Examining the choices we find our

Answer: first choice

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25mm of rain fell each second

Step-by-step explanation:

we know that 500 mm fell in 20 minutes

so, we have to divide 500 by 20 giving us the amount of rain that fell each minute:

500/20 = 25

therefore, 25mm of rain fell each minute

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Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne
Pavlova-9 [17]

Answer:

t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507  

p_v =P(t_{(15)}>1.507)=0.076

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

\bar X_{CTS}=2.35 represent the mean for the sample CTS

\bar X_{N}=1.83 represent the mean for the sample Normal

s_{CTS}=0.88 represent the sample standard deviation for the sample of CTS

s_{N}=0.54 represent the sample standard deviation for the sample of Normal

n_{CTS}=10 sample size selected for the CTS

n_{N}=7 sample size selected for the Normal

\alpha=0.01 represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis:\mu_{CTS} \leq \mu_{N}

Alternative hypothesis:\mu_{CTS} > \mu_{N}

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_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}} (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) the info given like this:

t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507  

P-value

The first step is calculate the degrees of freedom, on this case:

df=n_{CTS}+n_{N}-2=10+7-2=15

Since is a one side right tailed test the p value would be:

p_v =P(t_{(15)}>1.507)=0.076

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

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i saw another brainly and he seemed to be correct and i got it right :)!

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Answer:

3(m-4)=33

3m-12=33

3m=45

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Check:

3(15-4)=33

3(11)=33

33=33

Hope This Helps!!!

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