Answer:
And in order to obtain the confidence interval for the deviation we just take the square root and we got:
Since the confidence interval cointains the 1 we don't have enough evidence to reject the hypothesis given by the claim
Step-by-step explanation:
Data provided
1.9, 2.4, 3.0, 3.5, and 4.2
We can calculate the sample mean and deviation from this data with these formulas:
![\bar X = \frac{\sum_{i=1}^n X_i}{n}](https://tex.z-dn.net/?f=%5Cbar%20X%20%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20X_i%7D%7Bn%7D)
![s=\frac{\sum_{i=1}^n (X_i-\bar X)^2}{n-1}](https://tex.z-dn.net/?f=%20s%3D%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20%28X_i-%5Cbar%20X%29%5E2%7D%7Bn-1%7D)
And we got:
![\bar X= 3](https://tex.z-dn.net/?f=%5Cbar%20X%3D%203)
s=0.903 represent the sample standard deviation
n=5 the sample size
Confidence=95% or 0.95
Confidence interval
We need to begin finding the confidence interval for the population variance is given by:
The degrees of freedom given by:
The Confidence level provided is 0.95 or 95%, the significance is then
and
, and the critical values for this case are:
And the confidence interval would be:
And in order to obtain the confidence interval for the deviation we just take the square root and we got:
Since the confidence interval cointains the 1 we don't have enough evidence to reject the hypothesis given by the claim