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emmasim [6.3K]
4 years ago
14

Fill in the blank. For data sets having a distribution that is approximately​ bell-shaped, _______ states that about​ 68% of all

data values fall within one standard deviation from the mean
Mathematics
1 answer:
Harlamova29_29 [7]4 years ago
8 0

Answer:

Empirical Rule

Step-by-step explanation:

The Empirical Rule is also known as the Sigma rule or the 68-95-99.7% rule. The rule state that for a given data set, 68% of all data values will fall within the first standard deviation from the mean. The rule also states that 95% of all data values would fall within two standard deviations, while almost all the data which amounts to about 99.7% will fall within three standard deviations.

The empirical rule is used in forecasting based on the given datasets because it is a certainty that after obtaining the standard deviation, data values can be assigned to the categories they fall into, under the Empirical rule.

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In other way is

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Step-by-step explanation:

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A sample of 900900 computer chips revealed that 76v% of the chips do not fail in the first 10001000 hours of their use. The comp
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Answer:

Null hypothesis:p=0.78  

Alternative hypothesis:p \neq 0.78  

z=\frac{0.76 -0.78}{\sqrt{\frac{0.78(1-0.78)}{900}}}=-1.448  

p_v =2*P(Z  

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we see that 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 chip that do not fail in the first 1000 hours is not significantly different from 0.78 or 78%.  

Step-by-step explanation:

1) Data given and notation

n=900 represent the random sample taken

\hat p=0.76 estimated proportion of chips do not fail in the first 1000 hours

p_o=0.78 is the value that we want to test

\alpha represent the significance level

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 true proportion is 0.78:  

Null hypothesis:p=0.78  

Alternative hypothesis:p \neq 0.78  

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{0.76 -0.78}{\sqrt{\frac{0.78(1-0.78)}{900}}}=-1.448  

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 assumed for this case is \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(Z  

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we see that 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 chip that do not fail in the first 1000 hours is not significantly different from 0.78 or 78%.  

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Step-by-step explanation:

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