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stich3 [128]
3 years ago
8

A quality-control inspector is testing a batch of printed circuit boards to see wheater they are capable of performing in a high

temperature environment. He knows that the boards that will survive will pass all five of the tests with probability 98%. They will pass at least four tests with probability 99%, and they always pass at least three. On the other hand, the boards that will not survive sometimes pass the tests as well. In fact, 2% pass all five tests, and another 20% pass exactly four. The rest pass at most three tests. The inspector decides that if a board passes all five tests, he will classify it as "good." Otherwise, he'll classify it as "bad." The manager says that the probability of a type I error must be no larger than 0.01.
a. What does a type II error mean here?
b. What is the probability of a type II error?
Mathematics
1 answer:
avanturin [10]3 years ago
6 0

Answer:

see explaination

Step-by-step explanation:

Here the null hypothesis is that the PCB survives against the alternate that the PCB 'does not survive'. The test says that the PCB will survice if it is classified as 'good'; or, it will not survive if it is classifies as 'bad'.

a. The Type II error is the error committed when a PCB which cannot actually survive is classified as 'good'.

b. Therefore P(Type II error) = P(The PCB is classified as 'good' | PCB does not survives) = 0.03.

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Suppose that the researchers wanted to estimate the mean reaction time to within 6 msec with 95% confidence. Using the sample st
Lisa [10]

Answer:

a) The 95% confidence interval would be given by (509.592;550.308)  

b) n=523 rounded up to the nearest integer  

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X=530 represent the sample mean for the sample  

\mu population mean

s=70 represent the sample standard deviation  

n=48 represent the sample size (variable of interest)  

Confidence =95% or 0.995

Part a

The confidence interval for the mean is given by the following formula:  

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

The degrees of freedom are df=n-1=48-1=47

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,47)".And we see that z_{\alpha/2}=2.01  

Now we have everything in order to replace into formula (1):  

530-2.01\frac{70}{\sqrt{48}}=509.692  

530+2.01\frac{70}{\sqrt{48}}=550.308  

So on this case the 95% confidence interval would be given by (509.592;550.308)  

Part b

The margin of error is given by this formula:  

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (a)  

Assuming that \hat \sigma =s

And on this case we have that ME =6msec, and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=(\frac{z_{\alpha/2} \sigma}{ME})^2 (b)  

The critical value for 95% of confidence interval is provided, z_{\alpha/2}=1.96, replacing into formula (b) we got:  

n=(\frac{1.96(70)}{6})^2 =522.88 \approx 523  

So the answer for this case would be n=523 rounded up to the nearest integer  

3 0
2 years ago
The data in the table represent the training times (in seconds) for Adam. Adam 103 105 104 106 100 98 92 91 97 101 Points Concep
Julli [10]

Answer:

A. Miguel has the greatest spread.

The range of the two persons can be determined by:

Adam's range = 106 -91 = 15

Miguel's range = 105 -86 = 19

B. Considering the middle 50% of the training time, the person with the least spread is Adam.

Adam - 103 105 104 106 100

Miguel - 88 86 89 93 105

Adam's 50% range = 106 - 100 = 6

Miguel's 50% range = 105 - 86 = 19

Adam has the least spread

C. Miguel is inconsistent with the time set for training compared to that of Adam.

The answers to parts 2a and 2b show that there is a wide variation in the time that Miguel spend during training, but a minimum variation in the time spent by Adam during training.

3 0
3 years ago
Can someone help me please?
Alina [70]

Answer:

64/100 or 0.64 :)

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
What is the quotient of ÷
Svetlanka [38]
There isn't a quotient of the division sign
6 0
3 years ago
SoIve: 12 = -4(-6x - 3) <br><br><br> x = -0.625<br> x = -0.375<br> x = -2<br> x = 0
Vinvika [58]

Answer:

The answer is<u> "x = 0"</u>

<em>Step-by-step explanation:</em>

<em>Step-by-step explanation: </em><em>H</em><em>ope this answer is helpful</em><em>.</em><em>.</em><em>.</em>

<em> </em><em>Make me as brainliest...</em>

3 0
3 years ago
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