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Scilla [17]
2 years ago
14

Students are selling candy bars to raise money for a class trip. The number of candy bars sold image, and the amount of money ra

ised image, in dollars, are shown in the table below.
B E
5 $6.50
8 $10.40

Which equation represents the relationship between image and image?

A.
e=0.77b

B.
e = 1.3b

C.
e = 3.9b

D.
-
e = 1.73b
Mathematics
1 answer:
storchak [24]2 years ago
8 0

Answer:

B

Step-by-step explanation:

Because you divide b to e

1.3

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Answer:

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Step-by-step explanation:

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3 years ago
Suppose that 500 parts are tested in manufacturing and 10 are rejected.
alexdok [17]

Answer:

a) z=\frac{0.02 -0.03}{\sqrt{\frac{0.03(1-0.03)}{500}}}=-1.31  

p_v =P(Z  

If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of rejected items is less than 0.03.  

b) We can use a 95 percent upper confidence interval.

On this case we want a interval on this form : (-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})

So the critical value would be on this case Z_{\alpha}=1.64 and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"

We found the upper limit like this:

0.02+1.64\sqrt{\frac{0.02 (1-0.02)}{500}}=0.03026

And the interval would be: (-\infty,0.03026)

And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03

Step-by-step explanation:

Part a

Data given and notation  

n=500 represent the random sample taken

X=10 represent the number of objects rejected

\hat p=\frac{10}{500}=0.02 estimated proportion of objects rejected

p_o=0.03 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)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that 70% of adults say that it is morally wrong to not report all income on tax returns.:  

Null hypothesis:p=0.03  

Alternative hypothesis:p < 0.03  

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 requires we can replace in formula (1) like this:  

z=\frac{0.02 -0.03}{\sqrt{\frac{0.03(1-0.03)}{500}}}=-1.31  

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 one tailed left test the p value would be:  

p_v =P(Z  

If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of rejected items is less than 0.03.  

Part b

We can use a 95 percent upper confidence interval.

On this case we want a interval on this form : (-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})

So the critical value would be on this case Z_{\alpha}=1.64 and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"

We found the upper limit like this:

0.02+1.64\sqrt{\frac{0.02 (1-0.02)}{500}}=0.03026

And the interval would be: (-\infty,0.03026)

And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03

3 0
3 years ago
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