Mathematics, 10.12.2019 03:31, xxcecccc

Aspectrophotometer used for measuring co concentration [ppm (parts per million) by volume] is checked for accuracy by taking readings on a manufactured gas (called span gas) in which the co concentration is very precisely controlled at 70 ppm. if the readings suggest that the spectrophotometer is not working properly, it will have to be recalibrated. assume that if it is properly calibrated, measured concentration for span gas samples is normally distributed. on the basis of the six readings—69, 85, 72, 68, 77, and 82—is recalibration necessary? carry out a test of the relevant hypotheses using alpha = .05.
a) use the critical value approach
b) use the p-value approach

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a) The critical region is $t_{5}$ or $t_{5}2.57$ . If we analyze our calculated value 1.923 is not on the rejection zone of the null hypothesis. So we can't conclude that we have a significant difference for the value of 70ppm for the true mean.

b) $p_v =2*P(t_{(5)}1.923)=0.1125$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis. So we can't conclude that we have a significant difference for the value of 70ppm for the true mean.

Step-by-step explanation:

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is actually 70 or no, the system of hypothesis would be:

Null hypothesis: $\mu = 70$

Alternative hypothesis: $\mu \neq 70$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

First we need to calculate the sample mean and deviation given by:

$\bar x =\frac{\sum_{i=1}^n x_i}{n}=75.5$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar x)}{n-1}}=7.007$

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{75.5-70}{\frac{7.007}{\sqrt{6}}}=1.923$

a) Use the critical value approach

The first step is calculate the degrees of freedom, on this case:

df=n-1=6-1=5

Since we have a bilateral test we have two critical values, we need to look for a value that accumulates on each tail $\alpha/2=0.025$ of the area on the t distribution with 5 degrees of freedom. We can use the following excel codes to find it:

"=T.INV(0.025,5)" ,"=T.INV(1-0.025;5)"

And we got:

$z_{\alpha/2}=-2.57 , z_{1-\alpha/2}=2.57$

So our critical region is $t_{5}$ or $t_{5}2.57$ . If we analyze our calculated value 1.923 is not on the rejection zone of the null hypothesis. So we can't conclude that we have a significant difference for the value of 70ppm for the true mean.

b) Use the p-value approach

Since is a bilatral test the p value would be:

$p_v =2*P(t_{(5)}1.923)=0.1125$

answered: Guest

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2. incorrect.