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Leokris [45]
2 years ago
11

A group of screwdrivers produced in a day at a particular factory has a defect rate of 0.7%. Suppose a quality inspector randoml

y inspects 500 screwdrivers. Complete the following statement: The quality inspector should expect ____defective screwdrivers, with an error of ____. Group of answer choices
Mathematics
1 answer:
dangina [55]2 years ago
3 0

Answer: 3.5;0.4%

Step-by-step explanation:

The number of defective screwdrivers that the quality inspector should expect would be calculated as:

= np

= 0.7% × 500

= 0.7/100 × 500

= 0.007 × 500

= 3.5

.

The error will be calculated as:

= √p(1-p)/n

= ✓0.007(1 - 0.007)/500

= 0.004

= 0.4%

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1 year ago
According to a Washington Post-ABC News poll, 331 of 502 randomly selected U.S. adults interviewed said they would not be bother
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Answer:

z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124  

p_v =P(z>7.124)=5.24x10^{-13}  

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 interest is significantly higher than 0.5.  

Step-by-step explanation:

Data given and notation

n=502 represent the random sample taken

X=331 represent the adults that said they would not be bothered if the NAtional security agency

\hat p=\frac{331}{502}=0.659 estimated proportion of people who would not be bothered if the NAtional security agency

p_o=0.5 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

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 the true proportion is higher than the majority of 0.5:  

Null hypothesis:p\leq 0.5  

Alternative hypothesis:p > 0.5  

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{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124  

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=0.01. 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>7.124)=5.24x10^{-13}  

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 interest is significantly higher than 0.5.  

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Answer:

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38/6 = 6 remainder 2

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