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chubhunter [2.5K]
3 years ago
12

In a certain city, 27% of the families have a MasterCard, 19% have an American Express card, and 22% have a Visa card. Nine perc

ent of the families have both a MasterCard and an American Express card. Fourteen percent have both a Visa card and a MasterCard. Six percent have both an American Express card and a Visa card. What is the probability of selecting a family that has either a Visa card or an American Express card?
Mathematics
1 answer:
Natasha2012 [34]3 years ago
5 0

Answer:

35%

Step-by-step explanation:

Data provided in the question:

P(Master card) = 27% = 0.27

P(American express card) = 19% = 0.19

P( Visa card ) = 22% = 0.22

P( Master and American express card ) = 9% = 0.09

P( Visa and master card ) = 14% = 0.14

P( American express and visa card ) = 6% = 0.06

Now,

The  probability of selecting a family that has either a Visa card or an American Express card

= P( Visa card ) + P( American express card ) - P( Visa and American express card )

= 0.22 + 0.19 - 0.06

= 0.35

or

= 0.35 × 100% = 35%

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The mayor of a town has proposed a plan for the annexation of an adjoining community. A political study took a sample of 1000 vo
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Answer:

z=\frac{0.54 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.530  

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Step-by-step explanation:

Data given and notation

n=1000 represent the random sample taken

\hat p=0.54 estimated proportion of residents that favored the annexation

p_o=0.5 is the value that we want to test

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:  

Null hypothesis:p \leq 0.5  

Alternative hypothesis:p > 0.5  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info required we can replace in formula (1) like this:  

z=\frac{0.54 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.530  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

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

p_v =P(z>2.530)=0.0057  

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