Question

A statistical program is recommended. A spectrophotometer 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 69 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—77, 82, 72, 68, 69, and 85—is recalibration necessary? Carry out a test of the relevant hypotheses using α = 0.05. State the appropriate null and alternative hypotheses.

Answer 1

Answer:

$t=\frac{75.5-69}{\frac{7.007}{\sqrt{6}}}=2.272$  

$p_v =2*P(t_{(5)}>2.272)=0.072$  

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.  

We can say that at 5% of significance the true mean for the Co concentracion it's not significant different from 69.

Step-by-step explanation:

Data given and notation  

Data: 77, 82, 72, 68, 69, 85

The mean and sample deviation can be calculated from the following formulas:

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

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

$\bar X=75.5$ represent the sample mean  

$s=7.007$ represent the sample standard deviation  

$n=6$ sample size  

$\mu_o =69$ represent the value that we want to test  

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the population mean is different from 69, the system of hypothesis are :  

Null hypothesis:$\mu = 69$  

Alternative hypothesis:$\mu \neq 69$  

Since we don't know the population deviation, 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".  

Calculate the statistic  

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

$t=\frac{75.5-69}{\frac{7.007}{\sqrt{6}}}=2.272$  

P-value  

We need to calculate the degrees of freedom first given by:  

$df=n-1=6-1=5$  

Since is a two tailed test the p value would given by:  

$p_v =2*P(t_{(5)}>2.272)=0.072$  

Conclusion  

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.  

We can say that at 5% of significance the true mean for the Co concentracion it's not significant different from 69.