Final answer:
To test the claim that the sample mean head injury condition (hic) is less than 1000 at a 0.01 significance level, we can perform a one-sample t-test. The t-statistic is calculated using the sample mean, sample standard deviation, and the hypothesized mean. By comparing the t-statistic to the critical t-value, we can determine whether to reject or fail to reject the null hypothesis.
Step-by-step explanation:
To test the claim that the sample mean hic (head injury condition) is less than 1000 at a 0.01 significance level, we can perform a one-sample t-test. First, we calculate the sample mean of the data, which is (774 + 649 + 1210 + 546 + 431 + 612) / 6 = 753.67. Next, we calculate the sample standard deviation, which is the square root of the variance. The variance is equal to the sum of the squared differences between each data point and the mean, divided by (n-1), where n is the number of data points. Performing the calculations, we have: variance = ((774-753.67)^2 + (649-753.67)^2 + (1210-753.67)^2 + (546-753.67)^2 + (431-753.67)^2 + (612-753.67)^2) / (6-1) = 1201266.67 / 5 = 240253.33, and standard deviation = sqrt(variance) = sqrt(240253.33) = 490.13.
Using these values, we can calculate the t-statistic, which is equal to (sample mean - hypothesized mean) / (standard deviation / sqrt(n)), where n is the number of data points. In this case, the hypothesized mean is 1000, so the t-statistic = (753.67 - 1000) / (490.13 / sqrt(6)) = -2.748
Finally, we can compare the t-statistic to the critical t-value to determine whether to reject or fail to reject the null hypothesis. At a 0.01 significance level and degrees of freedom equal to (n-1), which in this case is 5, the critical t-value is -2.571. Since the t-statistic (-2.748) is less than the critical t-value (-2.571), we can reject the null hypothesis and conclude that the seats do not meet the safety requirement. Therefore, the correct answer is c) Reject the claim; the seats do not meet the safety requirement.