Question

A company produces rechargeable batteries for small size string trimmers. We define the life ofthe battery as the time between starting the string trimmer until the trimmer stops due to batteryfailure. It is known that the life of such batteries followsthe normal distribution with a knownpopulation standard deviation of 2.91 minutes. The company claims that the life of the newbattery that they have just produced is more than 30 minutes. To test the company’s claim, arandom sample of 25 batteries wereselected. Each battery was used in the same string trimmeruntil the trimmer stopped due to battery failure. In order to exclude the effect of the trimmer eachrun of the experiment was done on a different day. The sample mean life of the batteries was 31.1minutes. At the 5% level of significance, do the data indicate that the company’s claim is correct? Compute the sample test statistic and find the p-value.

Answer 1

One-Sample Z test

The sample mean is Xˉ=31.1, the population standard deviation is σ=2.91, and the sample size is n=25. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ =30 Ha: μ >30 This corresponds to a Right-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2a) Critical Value Based on the information provided, the significance level is α=0.05, therefore the critical value for this Right-tailed test is Zc​=1.6449. This can be found by either using excel or the Z distribution table. (2b) Rejection Region The rejection region for this Right-tailed test is Z>1.6449 (3) Test Statistics The z-statistic is computed as follows: (4) 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 p =P(Z>1.89)=0.0294 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that Z=1.89 > Zc​=1.6449, 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=0.0294, and since p=0.0294≤0.05, it is concluded that the null hypothesis is rejected. (6) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ  is greater than 30, at the 0.05 significance level.

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