Question

Imagine a cell phone manufacturer wants to test the claim that their cell phone batteries last more than 30 hours. Using a simple random sample of 18 batteries yielded a mean of 34.9 hours, with a standard deviation of 8.9 hours. Test this claim using a significance level of 0.05.

Answer 1

To test the claim that the cell phone batteries last more than 30 hours, we can use a hypothesis test with a significance level of 0.05. We calculate the test statistic and compare it to the critical value from the t-distribution. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that the batteries do last more than 30 hours.

Step-by-step explanation:

To test the claim that the cell phone batteries last more than 30 hours, we can use a hypothesis test. The null hypothesis is that the mean battery life is 30 hours, and the alternative hypothesis is that the mean battery life is greater than 30 hours. With a significance level of 0.05, we can calculate the test statistic and compare it to the critical value from the t-distribution. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that the batteries do last more than 30 hours.

First, let's calculate the test statistic:

t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))

t = (34.9 - 30) / (8.9 / sqrt(18))

t = 4.9 / (8.9 / 4.24264)

t = 4.9 / 2.09624

t ≈ 2.3345

Next, we need to find the critical value for a one-tailed test with a significance level of 0.05 and a degrees of freedom of 17 (n-1). From the t-distribution table or a calculator, the critical value is approximately 1.7408.

Since the test statistic (2.3345) is greater than the critical value (1.7408), we reject the null hypothesis. Therefore, we have enough evidence to support the claim that the cell phone batteries last more than 30 hours.