Mathematics, 30.11.2019 02:31 gardinerr410

An industrial engineer evaluating logic wafers was interested in the amount of silver required in the photographic stages. for 15 batches, the following net usages (ounces per wafer) were determined: 0.473 0.315 0.521 0.253 0.588 0.343 0.439 0.392 0.217 0.111 0.394 0.251 0.331 0.101 0.448 construct a 90% confidence interval for the mean silver usage per wafer. lower confidence bound= upper confidence bound=

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Answer from: potternatalie90

The 90% confidence interval is given by (0.281;0.409)

Lower Confidence bound =0.281

Upper Confidence Bound=0.409

Step-by-step explanation:

1) Notation and definitions

n=15 represent the sample size

$\bar X$ represent the sample mean

represent the sample standard deviation

m represent the margin of error

Confidence =90% or 0.90

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".

2) Calculate the mean and standard deviation for the sample

On this case we need to find the sample standard deviation with the following formula:

$s=\sqrt{\frac{\sum_{i=1}^{15}(x_i -\bar x)^2}{n-1}}$

And in order to find the sample mean we just need to use this formula:

$\bar x =\frac{\sum_{i=1}^{15} x_i}{n}$

The sample mean obtained on this case is $\bar x= 0.345$ and the deviation s=0.141

3) Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . The degrees of freedom are given by:

df=n-1=15-1=14

We can find the critical values in excel using the following formulas:

"=T.INV(0.05,14)" for $t_{\alpha/2}=-1.76$

"=T.INV(1-0.05,14)" for $t_{1-\alpha/2}=1.76$

The critical value $tc=\pm 1.76$

3) Calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

$m=t_c \frac{s}{\sqrt{n}}$

$m=1.76 \frac{0.141}{\sqrt{15}}=0.0642$

4) Calculate the confidence interval

The interval for the mean is given by this formula:

$\bar X \pm t_{c} \frac{s}{\sqrt{n}}$

And calculating the limits we got:

$0.345 - 1.76 \frac{0.141}{\sqrt{15}}=0.281$

$0.345 + 1.76 \frac{0.141}{\sqrt{15}}=0.409$

The 90% confidence interval is given by (0.281;0.409)

Answer from: Quest

c. 4 and 1/2 pounds is the answer! : )

Answer from: Quest

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.

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An industrial engineer evaluating logic wafers was interested in the amount of silver required in th...