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Pavlova-9 [17]
2 years ago
5

Juan has 5 counters. Carlos has 60 counters.

Mathematics
2 answers:
Oksana_A [137]2 years ago
7 0

Answer:

12 counters

Step-by-step explanation:

Since Juan has 5 counters and 5×12=60. CarLos has 60 of them.

alekssr [168]2 years ago
3 0

Answer: 12

Step-by-step explanation: 60/5=12

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Answer:

Null hypothesis:p_{1} = p_{2}  

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

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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  

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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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