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RUDIKE [14]
3 years ago
10

A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is a black card given that it is an

ace.
Mathematics
1 answer:
Pavlova-9 [17]3 years ago
5 0

probability of getting black card an ace =2/52=1/26

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Answer:

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3 years ago
According to the University of Nevada Center for Logistics Management, of all merchandise sold in the United States gets returne
Step2247 [10]

Answer:

a) \hat p=\frac{12}{80}=0.15 estimated proportion of items that were returned

b) The 95% confidence interval would be given (0.0718;0.228).

c) Using a significance level assumed \alpha=0.05 we see that p_v so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of returns at the Houston store significantly different from the returnsfor the nation as a whole.  

Step-by-step explanation:

Assuming:

According to the University of Nevada Center for Logistics Management, 6% of all mer-chandise sold in the United States gets returned. Houston department store sampled 80 items sold in January and found that 12 of the items  were returned.

Data given and notation  

n=80 represent the random sample taken    

X=12 represent the items  that were returned

\hat p=\frac{12}{80}=0.15 estimated proportion of items that were returned

\alpha=0.05 represent the significance level (no given, but is assumed)    

Confidence =0.95 or 95%

p= population proportion of items  that were returned

a. Construct a point estimate of the proportion of items returned for the population ofsales transactions at the Houston store

\hat p=\frac{12}{80}=0.15 estimated proportion of items that were returned

b. Construct a 95% confidence interval for the porportion of returns at the Houston store

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

And replacing into the confidence interval formula we got:

0.15 - 1.96 \sqrt{\frac{0.15(1-0.15)}{80}}=0.0718

0.15 + 1.96 \sqrt{\frac{0.15(1-0.15)}{80}}=0.228

And the 95% confidence interval would be given (0.0718;0.228).

c. Is the proportion of returns at the Houston store significantly different from the returns for the nation as a whole? Provide statistical support for your answer.

We need to conduct a hypothesis in order to test the claim that the population proportion differs significantly to the USA proportion of 6% or no. We have the following system of hypothesis :    

Null Hypothesis: p = 0.06  

Alternative Hypothesis: p \neq 0.06  

We assume that the proportion follows a normal distribution.    

This is a two tailed test for the proportion .  

The One-Sample Proportion Test is "used to assess whether a population proportion \hat p is significantly (different,higher or less) from a hypothesized value p_o".  

Check for the assumptions that he sample must satisfy in order to apply the test  

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.  

b) The sample needs to be large enough  

np_o =80*0.15=12>10  

n(1-p_o)=80*(1-0.15)=68>10  

Calculate the statistic    

The statistic is calculated with the following formula:  

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

On this case the value of p_o=0.06 is the value that we are testing and n = 80.  

z=\frac{0.15 -0.06}{\sqrt{\frac{0.06(1-0.06)}{80}}}=3.390

The p value for the test would be:  

p_v =2*P(z>3.390)=0.00070

Using a significance level assumed \alpha=0.05 we see that p_v so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion of returns at the Houston store significantly different from the returnsfor the nation as a whole.  

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Step-by-step explanation: You will definitely never in your life need to know more than that :) I suggest you make a song to remember.

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