Question

An experiment was conducted to compare the lifetimes of two different car batteries(from two different manufacturer) that both claim a lifetime of at least 4 years. 12 batteries from apremium manufacturer were tested and the results were a mean lifetime of 4.31 years with a standarddeviation of 0.39 years. 15 batteries from a budget manufacturer were tested and the results werea mean lifetime of 3.94 years with a standard deviation of 0.63 years. Can you conclude up to aα= 0.05level of significance that the premium batteries last longer than the budget batteries? Thebatteries population standard deviations are unknown but assumed not to be equal.

Answer 1

T-test for two Means – Unknown Population Standard Deviations - Unequal Variance

The following information about the samples has been provided: a. Sample Means : Xˉ1​=4.31 and Xˉ2​=3.94 b. Sample Standard deviation: s1=0.39 and s2=0.63 c. Sample size: n1=12 and n2=15 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ1 =μ2 Ha: μ1 >μ2 This corresponds to a Right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) The degrees of freedom Assuming that the population variances are unequal, the degrees of freedom are given by (3a) Critical Value Based on the information provided, the significance level is α=0.05, and the degree of freedom is 23.7029. Therefore the critical value for this Right-tailed test is tc​=1.7139. This can be found by either using excel or the t distribution table. (3b) Rejection Region The rejection region for this Right-tailed test is t>1.7139 (4)Test Statistics The t-statistic is computed as follows: (5) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0.0371 (6) The Decision about the null hypothesis (a) Using traditional method Since it is observed that t=1.8703 > tc​=1.7139, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.0371, and since p=0.0371≤0.05, it is concluded that the null hypothesis is rejected. (7) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ1​ is greater than μ2, at the 0.05 significance level.

Let me know in the comments if anything is not clear. I will reply ASAP! Please do upvote if satisfied!