Final answer:
To test if we can infer at the 2% significance level that the population mean is not equal to 15, we can use a one-sample t-test. The standardized test statistic is -1.897. The rejection region is (-infinity, -2.821) ∪ (2.821, infinity) and the p-value is approximately 0.043.
Step-by-step explanation:
To test if we can infer at the 2% significance level that the population mean is not equal to 15, we can use a one-sample t-test since the population standard deviation is unknown. The null hypothesis (H0) is that the population mean is equal to 15, and the alternative hypothesis (Ha) is that the population mean is not equal to 15. To calculate the value of the standardized test statistic, we use the formula: t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size)). In this case, the values are: t = (12.8 - 15) / (2 / sqrt(10)) = -1.897. The standardized test statistic is -1.897. The rejection region for a two-tailed test at the 2% significance level is found by dividing the significance level by 2, so the rejection region is (-infinity, -tα/2) ∪ (tα/2, infinity). Using a t-table or calculator, the critical value for the 1% significance level with 9 degrees of freedom is approximately ±2.821. Therefore, the rejection region is (-infinity, -2.821) ∪ (2.821, infinity). The p-value is the probability of observing a test statistic as extreme or more extreme than the one obtained, assuming the null hypothesis is true. To calculate the p-value, we calculate the area under the t-distribution curve to the left of the test statistic and multiply it by 2 for a two-tailed test. In this case, the p-value is approximately 0.043.