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2 years ago

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/ for the fraction bar(s). The data set gives the number of bottles filled by each of the workers in a bottling plant in one day. {36, 18, 16, 28, 68, 35, 37, 66, 38, 40, 41, 44, 72, 29} The best measure of center for this data set is the , and its value expressed up to one decimal place is .

2 answers:

geniusboy [140]2 years ago

7 0

Answer:

The best measure of center for this data set is the median and its value expressed up to one decimal place is 37.50

Step-by-step explanation:

The best measure of center for this data set is the MEDIAN , and its value expressed up to one decimal place is 37.50.

We have given the data set gives the number of bottles filled by each of the workers in a bottling plant in one day.

The data set is:

{36, 18, 16, 28, 68, 35, 37, 66, 38, 40, 41, 44, 72, 29}

Let us find the median of this data set.

Arrange the values in ascending order:

{16,18,28,29,35,36,37,38,40,41,44,66,68,72}

Median = average of middle entries

Median = 37.5

Therefore the best measure of center for this data set is the median and its value expressed up to one decimal place is 37.50 ....

choli [55]2 years ago

6 0

Answer:

The best measure of center for this data set is the mean and its value expressed up to one decimal place is 40.6

Step-by-step explanation:

The best measure of center for this data set is the mean.

Calculating Mean......

Mean is calculated as summation of observation divided by number of observations.

Given the following dataset;

36, 18, 16, 28, 68, 35, 37, 66, 38, 40, 41, 44, 72, 29

The sum = 568

Number of observations (data set) = 14

Mean = 568/14

Mean = 40.571428571

Mean = 40.6 --- Approximated

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Null hypothesis: $p_{1} = p_{2}$

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

$z=\frac{0.179-0.15}{\sqrt{0.17(1-0.17)(\frac{1}{140}+\frac{1}{60})}}=0.500$

$p_v =2*P(Z>0.500)=0.617$

So the p value is a very low 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 we can say the two proportions NOT differs significantly.

Step-by-step explanation:

Data given and notation

$X_{1}=25$ represent the number of homeowners who would buy the security system

$X_{2}=9$ represent the number of renters who would buy the security system

$n_{1}=140$ sample 1

$n_{2}=60$ sample 2

$p_{1}=\frac{25}{140}=0.179$ represent the proportion of homeowners who would buy the security system

$p_{2}=\frac{9}{60}= 0.15$ represent the proportion of renters who would buy the security system

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the two proportions differs , the system of hypothesis would be:

Null hypothesis: $p_{1} = p_{2}$

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

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 $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{25+9}{140+60}=0.17$

Calculate the statistic

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

$z=\frac{0.179-0.15}{\sqrt{0.17(1-0.17)(\frac{1}{140}+\frac{1}{60})}}=0.500$

Statistical decision

For this case we don't have a significance level provided $\alpha$ , but we can calculate the p value for this test.

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

$p_v =2*P(Z>0.500)=0.617$

So the p value is a very low 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 we can say the two proportions NOT differs significantly.

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