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

Answer:

$\frac{(4)(0.903)^2}{11.143} \leq \sigma^2 \leq \frac{(4)(0.903)^2}{0.484}$

$0.293 \leq \sigma^2 \leq 6.736$

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

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:

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

$s=\frac{\sum_{i=1}^n (X_i-\bar X)^2}{n-1}$

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:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

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:

$\frac{(4)(0.903)^2}{11.143} \leq \sigma^2 \leq \frac{(4)(0.903)^2}{0.484}$

$0.293 \leq \sigma^2 \leq 6.736$

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