Question

A consumer organization collected data on two types of automobile batteries, A and B. The summary statistics for 12 observations of each type are . X(bar)A=36.51, X(bar)B=34.21, SA=1.43, SB=.93. Assume that the data are normally distributed with OA=OB.

Is there evidence to support the claim that type A battery mean life exceeds that of type B? Test the hypothesis at alpha = 0.05

Construct a 99% Confidence interval (CI) for the difference in the mean battery life.

Answer 1

Answer:

There evidence to support the claim that type A battery mean life exceeds that of type B.

The 99% Confidence interval (CI) for the difference in the mean battery life is (0.91, 3.69).

Step-by-step explanation:

The hypothesis can be defined as follows:

H₀:The mean life of battery A is not more than that of battery B, i.e. $\mu_{A}-\mu_{B}\leq 0$.

Hₐ:The mean life of battery A is more than that of battery B, i.e. $\mu_{A}-\mu_{B}>0$.

The significance level of the test is, α = 0.05.

The test statistic is:

$t=\frac{\bar x_{A}-\bar x_{B}}{SE}$

Compute the value of standard error as follows:

$SE=\sqrt{\frac{s_{A}^{2}}{n_{A}}+\frac{s_{B}^{2}}{n_{B}}}=\sqrt{\frac{1.43^{2}}{12}+\frac{0.93^{2}}{12}}=\sqrt{0.242483}=0.4924$

Compute the test statistic as follows:

$t=\frac{\bar x_{A}-\bar x_{B}}{SE}=\frac{36.51-34.21}{0.4924}=4.67$

Decision rule:

If the test statistic value is more than the critical value, $t_{\alpha, df}$, then the null hypothesis will be rejected. And vice-versa.

Compute the degrees of freedom (df) as follows:

$df=\frac{SE^{4}}{\frac{(s_{A}^{2}/n_{A})^{2}}{n_{A}-1}+\frac{(s_{B}^{2}/n_{B})^{2}}{n_{B}-1}}=\frac{0.4924^{4}}{0.022}=18.88\approx19$

The critical value is, $t_{\alpha, df}=t_{0.05, 19}=1.729$

The test statistic, t = 4.67 > $t_{\alpha, df}$ = 1.729.

Thus, the null hypothesis is rejected at 5% level of significance.

Conclusion:

Thus, there evidence to support the claim that type A battery mean life exceeds that of type B.

The critical value of t for a 99% confidence level and degrees of freedom 19 is:

$t_{0.01/2, 22}=2.819$

The pooled standard deviation is:

$S_{p}^{2}=\frac{(n_{A}-1)s_{A}^{2}+(n_{B}-1)s_{B}^{2}}{n_{A}+n_{B}-2}=\frac{(11\times1.43^{2})+(11\times0.93^{2})}{22}=1.455$

Compute the 99% confidence interval for the difference in the mean battery life as follows:

$CI=\bar x_{A}-\bar x_{B}\pm t_{00.01/2, 22}\times \sqrt{S_{p}^{2}[\frac{1}{n_{A}}+\frac{1}{n_{B}}]}\\=(36.51-34.21)\pm 2.819\times\sqrt{1.455[\frac{1}{12}+\frac{1}{12}]}\\=2.3\pm1.39\\=(0.91, 3.69)$

Thus, the 99% Confidence interval (CI) for the difference in the mean battery life is (0.91, 3.69).