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Rzqust [24]
2 years ago
13

The type of part being manufactured today should be 1.27 inches wide, with a tolerance of 0.04 inches. If a part does not meet t

he size requirement, it must be rejected. You have 4 parts to inspect. They have the following widths. How many must be rejected? 1.275 inches, 1.300 inches, 1.266 inches, 1.255 inches A. None B. 1 C. 2 D. 3 E. 4
Mathematics
2 answers:
nydimaria [60]2 years ago
8 0

Answer:

option: A is correct.

None of the parts will get rejected.

Step-by-step explanation:

The type of part being manufactured today should be 1.27 inches wide, with a tolerance of 0.04 inches.

i.e. the range upto which the manufactured part won't be rejected are:

lower value:

1.27-0.04=1.23

and 1.27+0.04=1.31

So, any part whose width is between:

1.23 inches - 1.31 inches won't get rejected.

1)

1.275 inches.

as 1.275 inches lies between 1.23-1.31 inches.

Hence it won't get rejected.

2)

1.300 inches.

As it also lies between 1.23-1.31 inches.

Hence it won't get rejected.

3)

1.266 inches.

As it also lies between 1.23-1.31 inches.

Hence it won't get rejected.

4)

1.255 inches

As it also lies between 1.23-1.31 inches.

Hence it won't get rejected.

This means that none of the part will get rejected.

olasank [31]2 years ago
3 0
The range of widths is {1.23,1.31} so none will be rejected, answer A.

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Nataly_w [17]

Answer:

The p value obtained and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidenc to reject the null hypothesis, and we can said that at 5% of significance the proportion of new laptop's with fully charge for the new model is significantly higher compared to the old model.  

Step-by-step explanation:

1) Data given and notation n  

n=100 represent the random sample taken

X=96 represent the laptop's arrived with fully charged batteries.

\hat p=\frac{96}{100}=0.96 estimated proportion of laptop's arrived with fully charged batteries.

p_o=0.85 is the value that we want to test

\alpha=0.05 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 new model’s rate is at least as high as the previous model.:  

Null hypothesis:p \leq 0.85  

Alternative hypothesis:p > 0.85  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

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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:  

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4) Statistical decision  

P value method or p value approach . "This method consists on 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 is not provided, but we can assume \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a one side test the p value would be:  

p_v =P(z>3.08)=0.0010  

So based on the p value obtained and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidenc to reject the null hypothesis, and we can said that at 5% of significance the proportion of new laptop's with fully charge for the new model is significantly higher compared to the old model.  

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