Mathematics, 07.03.2020 03:36 brinicole1011

The following sample of weights was taken from 8 muffins off the assembly line. Construct the 90% confidence interval for the population variance for all muffins that come off the assembly line. Round your answers to two decimal places. 12.9,12.5,13.0,13.1,12.3,12.9,12.2, 13.2

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Answer from: manny2085

$\frac{(7)(0.378)^2}{14.067} \leq \sigma^2 \leq \frac{(7)(0.378)^2}{2.167}$

$0.0711 \leq \sigma^2 \leq 0.462$

And rounded we got:

$0.07 \leq \sigma^2 \leq 0.46$

Step-by-step explanation:

Data given and notation

s represent the sample standard deviation

$\bar x$ represent the sample mean

n=8 the sample size

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 mean or variance 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.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

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

$s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}$

The sample mean obtained on this case is $\bar x= 12.7625$ and the deviation s=0.378

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=8-1=7

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,7)" "=CHISQ.INV(0.95,7)". so for this case the critical values are:

$\chi^2_{\alpha/2}=14.067$

$\chi^2_{1- \alpha/2}=2.167$

And replacing into the formula for the interval we got:

$\frac{(7)(0.378)^2}{14.067} \leq \sigma^2 \leq \frac{(7)(0.378)^2}{2.167}$

$0.0711 \leq \sigma^2 \leq 0.462$

And rounded we got:

$0.07 \leq \sigma^2 \leq 0.46$

Answer from: Quest

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that runner b is moving at a slower rate."

Answer from: Quest

40 x 50 = 2000

step-by-step explanation:

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