is strictly increasing on [0, 5], so
and
so the integral is bounded between
[asy] size(7cm); draw((1.8,0)--(3.2,0),Arrows); draw((2,0.03)--(2,-0.03)); draw((3,0.03)--(3,-0.03)); for(real i=15/7; i<2.9;
1 answer:
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Answer:
Traditional method
We can find a critical value in the chi square distribution with df =16 who accumulates of the area on each tail and we got:
Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is NOT significantly different from 3.5 cm
Confidence level method
We can find the p value like this
Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is NOT significantly different from 3.5 cm
Step-by-step explanation:
Data provided
represent the sample selected
represent the significance
represent the sample variance
represent the value to check
Null and alternative hypothesis
We want to verify if the new production method has lengths with a standard deviation different from 3.5 cm, so the system of hypothesis would be:
Null Hypothesis:
Alternative hypothesis:
The statistic for this case is given by:
The degrees of freedom are:
Traditional method
We can find a critical value in the chi square distribution with df =16 who accumulates of the area on each tail and we got:
Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is significantly different from 3.5 cm
Confidence level method
We can find the p value like this
Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is significantly different from 3.5 cm