Question

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=

Answer 1

Answer:

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  

$s$ 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)