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ElenaW [278]
3 years ago
11

In 2001, a company marketed 730,000 units of its product. In 2001 its yearly volume was 50% of its volume for 2004. The 2004 vol

ume represents how many units for each of the 365 days of 2004?
Mathematics
2 answers:
borishaifa [10]3 years ago
7 0

Answer: 4,000 units per day.

Step-by-step explanation:

You know that in 2001 its yearly volume was 50% of its volume for 2004.

Therefore, if in 2001 the company marketed 730,000 units of its product, in 2004 the volume is:

Yearly\ volume_{(2004)}=(730,000\ units)(2)=1,460,000\ units

Let be "x" the number of units for each of the 365 days of 2004, you can find its value by dividing  the 2004 volume by 365.

The result is:

units=\frac{1,460,000\ units}{365}=4,000\ units.

kotykmax [81]3 years ago
6 0

Answer:

4000 units

Step-by-step explanation:

In 2001 the total number of marketed units= 730,000

If this number represents 50% of what was marketed in 2004, then the total number of units marketed in 2004 was:

(100/50)× 730,000=1460000

To get the number for each of the 365 days in 2004 we divide the total for 2004 by 365

1460000/365= 4000 units

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A TV station claims that 38% of the 6:00 - 7:00 pm viewing audience watches its evening news program. A consumer group believes
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Answer:

z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374  

p_v =P(Z  

So 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 people that regularly watch the TV station’s news program is significantly less than 0.38 .  

Step-by-step explanation:

1) Data given and notation

n=830 represent the random sample taken

X=282 represent the people that regularly watch the TV station’s news program

\hat p=\frac{282}{830}=0.340 estimated proportion of people that regularly watch the TV station’s news program

p_o=0.38 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 the proportion is less than 0.38:  

Null hypothesis:p\geq 0.38  

Alternative hypothesis:p < 0.38  

When we conduct a proportion test we need to use the z statisitc, 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{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374  

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 left tailed test the p value would be:  

p_v =P(Z  

So 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 people that regularly watch the TV station’s news program is significantly less than 0.38 .  

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