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Juliette [100K]
3 years ago
9

A retail shop accepts only cash or checks. Suppose that 57% of its customers carry cash, 50% carry checks, and 39% carry both ca

sh and checks. What is the
probability that a randomly chosen customer at the shop is carrying cash or checks (or both)?
Write your answer as a decimal (not as a percentage).
Mathematics
1 answer:
Vanyuwa [196]3 years ago
5 0

A retail shop accepts only cash or checks. Suppose that

46

% of its customers carry cash,

37

% carry checks, and

74

% carry cash or checks (or both). What is the probability that a randomly chosen customer at the shop is carrying both cash and checks?

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4.One attorney claims that more than 25% of all the lawyers in Boston advertise for their business. A sample of 200 lawyers in B
AleksAgata [21]

Answer:

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

p_v =P(Z>2.123)=0.0169  

The p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

Step-by-step explanation:

1) Data given and notation  

n=200 represent the random sample taken

X=63 represent the lawyers had used some form of advertising for their business

\hat p=\frac{63}{200}=0.315 estimated proportion of lawyers had used some form of advertising for their business

p_o=0.25 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that more than 25% of all the lawyers in Boston advertise for their business:  

Null hypothesis:p\leq 0.25  

Alternative hypothesis:p > 0.25  

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.

3) Calculate the statistic  

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

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

4) 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 significance level provided \alpha=0.05. 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>2.123)=0.0169  

The p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

8 0
4 years ago
A computer supplies store is having a store wide sale this weekend. And inkjet printer that normally sells for $179 is on sale f
klasskru [66]
Hey again!

i don't know if i am doing this correctly but i will give it a go

179 - 143.20 = 35.8

179 / 35.8 = 5

so this would mean from 179 to 143.20 their is a 5% discount

Hope This Helps!
8 0
4 years ago
A clinical psychologist wants to test whether experiencing childhood trauma affects one's self-efficacy in adulthood. He randoml
Verdich [7]

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 152.5

For the alternative hypothesis,

µ ≠ 152.5

This is a two tailed test.

Since no population standard deviation is given, the distribution is a student's t.

Since n = 231

Degrees of freedom, df = n - 1 = 231 - 1 = 230

t = (x - µ)/(s/√n)

Where

x = sample mean = 148.9

µ = population mean = 152.5

s = samples standard deviation = 27.4

t = (148.9 - 152.5)/(27.4/√231) = - 2

We would determine the p value using the t test calculator. It becomes

p = 0.047

Since alpha, 0.05 > thanthere sufficient evidence to conclude that the self-efficacy of adults who have experienced childhood trauma differs from that in the general population of individuals the p value, 0.047, then we would reject the null hypothesis. Therefore, At a 5% level of significance, there is sufficient evidence to conclude that the self-efficacy of adults who have experienced childhood trauma differs from that in the general population of individuals

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3 years ago
3.7 ÷ 1.1 = ? Round to the nearest hundredth.
noname [10]
The answer is D) 3.36.
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<span>Simplify (8 + 7i) + (2 –i) 
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