Final answer:
A hypothesis test is performed to determine if there is sufficient evidence to support the principal's claim. The null hypothesis is that the mean IQ score of the students is equal to the population mean of 100, while the alternative hypothesis is that the mean IQ score is greater than 100. Using a significance level of 0.05, the test statistic is calculated and the p-value is found to be approximately 0.0002. Since the p-value is less than the significance level, the null hypothesis is rejected, indicating sufficient evidence to support the principal's claim.
Step-by-step explanation:
To determine if there is sufficient evidence to support the principal's claim, we can perform a hypothesis test. The null hypothesis, H0, would be that the mean IQ score of the students is equal to the population mean of 100. The alternative hypothesis, Ha, would be that the mean IQ score of the students is greater than 100.
Using a significance level of 0.05, we can calculate the test statistic and find the p-value. Since we have a sample size of 30, we can assume the sample mean follows a normal distribution.
Using the given information, the test statistic is calculated as follows:
t = (sample mean - population mean) / (standard deviation / sqrt(sample size)) = (112 - 100) / (15 / sqrt(30)) ≈ 4.08 (rounded to two decimal places)
Consulting a t-distribution table or using statistical software, we can find the p-value associated with this test statistic. The p-value is the probability of obtaining a test statistic as extreme as the calculated t-value, assuming the null hypothesis is true.
Upon calculation, the p-value is approximately 0.0002 (rounded to four decimal places). Since the p-value is less than the significance level of 0.05, we can reject the null hypothesis. Therefore, there is sufficient evidence to support the principal's claim that the students in his school have above average intelligence.