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2 years ago

If the linear regression for a calibration curve is y=1500x – 200 for x= 0.5 y= ?

Mathematics

1 answer:

Dafna11 [192]2 years ago

3 0

The value of y from the given equation if x is 0.5, is 450

<h3>What is an equation?</h3>

They are expressions separated by an equal sign

Given the linear regression for a calibration curve expressed as:

y=1500x – 200

If the value of x is 0.5, then;

y = 1500(0.5) - 200

y = 750 - 200

y = 550

Hence the value of y if x is 0.5, is 450

Learn more on linear regression here: brainly.com/question/25987747

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2 years ago

A sample size must be determined for estimating a population mean given that the confidence level is % and the desired margin o

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Complete Question

The complete question is shown on the first uploaded image

Answer:

$n =290$

$n =129$

The correct option is D

Step-by-step explanation:

From the question we are told that

The margin of error is $E = 0.23$

The largest value is k = 15

The smallest value is u = 7

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally population standard deviation estimate is mathematically represented as

$\sigma = \frac{Range }{4}$

Here the Range is mathematically represented as

$Range = k - u$

=> $Range = 15 - 7$

=> $Range = 8$

=> $\sigma = \frac{8 }{4}$

=> $\sigma = 2$

Generally the sample size is mathematically represented as

$n = [\frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ] ^2$

=> $n = [1.96 * 2 }{0.23} ] ^2$

=> $n =290$

Generally to obtain a conservatively small sample size the population standard deviation estimate becomes

=> $\sigma = \frac{Range }{6}$

=> $\sigma = \frac{8 }{6}$

=> $\sigma =1.333$

Generally the conservatively small sample size is mathematically evaluated as

$n = [\frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ] ^2$

=> $n = [1.96 * 1.333 }{0.23} ] ^2$

=> $n =129$

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2 years ago