Question

A manufacturer of car batteries claims that the batteries will last, on average, 3 years with a variance of 1 year. If 5 of these batteries have lifetimes of 1.9, 2.4, 3.0, 3.5, and 4.2 years, construct a 95% confidence interval for σ2 and decide if the manufacturer’s claim that σ2 = 1 is valid. Assume the population of battery lives to be approximately normally distributed.

Answer 1

`((4)(0.903)^2)/(11.143) \leq \sigma^2 \leq ((4)(0.903)^2)/(0.484)`

`0.293 \leq \sigma^2 \leq 6.736`

And in order to obtain the confidence interval for the deviation we just take the square root and we got:

`0.541 \leq \sigma \leq 2.595`

Since the confidence interval cointains the 1 we don't have enough evidence to reject the hypothesis given by the claim

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 = (\sum_(i=1)^n X_i)/(n)`

`s=(\sum_(i=1)^n (X_i-\bar X)^2)/(n-1)`

And we got:

`\bar X= 3`

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:

`((n-1)s^2)/(\chi^2_(\alpha/2)) \leq \sigma^2 \leq ((n-1)s^2)/(\chi^2_(1-\alpha/2))`

The degrees of freedom given by:

`df=n-1=5-1=4`

The Confidence level provided is 0.95 or 95%, the significance is then
`\alpha=0.05` and
`\alpha/2 =0.025`, and the critical values for this case are:

`\chi^2_(\alpha/2)=11.143`

`\chi^2_(1- \alpha/2)=0.484`

And the confidence interval would be:

`((4)(0.903)^2)/(11.143) \leq \sigma^2 \leq ((4)(0.903)^2)/(0.484)`

`0.293 \leq \sigma^2 \leq 6.736`

And in order to obtain the confidence interval for the deviation we just take the square root and we got:

`0.541 \leq \sigma \leq 2.595`

Since the confidence interval cointains the 1 we don't have enough evidence to reject the hypothesis given by the claim