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slega [8]
3 years ago
14

What is the Interquartile Range for the following data set? (5,6,7,3,9,8,3,1, 6, 7,7} IQR:

Mathematics
2 answers:
WITCHER [35]3 years ago
6 0
4 I hope this helped. Same answer as above tho
iren2701 [21]3 years ago
4 0

Answer:

IQR = 4

Step-by-step explanation:

The interquartile range is the difference between the upper quartile and the lower quartile.

The first step is to find the median of the data set.

The median is the middle value of the data arranged in ascending order.

Arranging the data in ascending order

1 , 3 , 3 , 5 , 6 , 6 , 7 , 7 , 7 , 8 , 9

                        ↑ median is 6

The upper quartile is the middle value to the right of the median

7 , 7 , 7 , 8 , 9

         ↑ upper quartile is 7

The lower quartile is the middle value to the left of the median

1 , 3 , 3 , 5 , 6

         ↑ lower quartile is 3

Thus

Interquartile range = 7 - 3 = 4

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Not Answered Country Financial, a financial services company, uses surveys of adults age 18 and older to determine if personal f
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Answer:

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}    

z=\frac{0.410-0.350}{\sqrt{0.382(1-0.382)(\frac{1}{1000}+\frac{1}{900})}}=2.688    

p_v =2*P(Z>2.688)= 0.0072    

Comparing the p value with the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportions analyzed are significantly different at 5% of significance.

Step-by-step explanation:

Data given and notation    

X_{1}=410 represent the number of people indicating that their financial security was more than fair  in 2012

X_{2}=315 represent the number of people indicating that their financial security was more than fair in 2010

n_{1}=1000 sample 1 selected  

n_{2}=900 sample 2 selected  

p_{1}=\frac{410}{1000}=0.410 represent the proportion estimated of people indicating that their financial security was more than fair in 2012

p_{2}=\frac{315}{900}=0.350 represent the proportion estimated of people indicating that their financial security was more than fair in 2010  

\hat p represent the pooled estimate of p

z would represent the statistic (variable of interest)    

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

\alpha=0.05 significance level given  

Part a: Concepts and formulas to use    

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:    

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}    

Hypothesis testing

We need to apply a z test to compare proportions, and the statistic is given by:    

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{410+315}{1000+900}=0.382  

z-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  

Replacing in formula (1) the values obtained we got this:    

z=\frac{0.410-0.350}{\sqrt{0.382(1-0.382)(\frac{1}{1000}+\frac{1}{900})}}=2.688    

Statistical decision  

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

p_v =2*P(Z>2.688)= 0.0072    

Comparing the p value with the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportions analyzed are significantly different at 5% of significance.

3 0
3 years ago
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