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sertanlavr [38]
2 years ago
12

A batch of 20 semiconductor chips is inspected by choosing a sample of 3 chips. Assume 10 of the chips do not conform to custome

r requirements. Round your answers to the nearest integer. a. How many different samples are possible? b. How many samples of 3 contain exactly one nonconforming chip? c. How many samples of 3 contain at least one nonconforming chip?
Mathematics
1 answer:
Hatshy [7]2 years ago
6 0

Answer:

A batch of 20 semiconductor chips is inspected by choosing a sample of 3 chips. Assume 10 of the chips do not conform to customer requirements.

a) Number of different samples = 20C3 =  \frac{20!}{3!(17!)} =1140

b) 2 good and one bad chip

Number of samples = 10C2 * 10C1 = 45 * 10 =450

c) 2 good 1 bad + 1 good 2 bad + 3 bad

Number of samples = 10C2 * 10C1 + 10C1 * 10C2 + 10C3

= 450+450+120=1020

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Ashley invests $9,720 in a one-month money market account paying 3.16% simple annual interest and $8,140 in a two-year CD yieldi
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For the first investment. A = P(1 + rt); where p = 9,720, r = 0.0316 and t = 1/12
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2 years ago
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Most US adults have social ties with a large number of people, including friends, family, co-workers, and other acquaintances. I
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Answer:

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

p_v =2*P(t_{(1699)}>1.971)=0.049  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation  

\bar X=669 represent the sample mean

s=732 represent the sample standard deviation

n=1700 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0. represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:  

Null hypothesis:\mu = 634  

Alternative hypothesis:\mu \neq 634  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=1700-1=1699  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(1699)}>1.971)=0.049  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

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