Final answer:
The p-value associated with a z-test statistic of 2.98 at a 0.01 significance level is 0.0014, indicating that we would reject the null hypothesis that exactly 50% of senior executives believe the most common interview mistake is little or no knowledge of the company.
Step-by-step explanation:
The question involves determining the p-value from a given z-test statistic to assess the validity of a claim made based on survey results. The null hypothesis in this scenario is that 50% of senior executives say that the most common job interview mistake is to have little or no knowledge of the company. With a test statistic of z = 2.98 and a significance level (α) of 0.01, we need to find the probability that a standard normal distribution is greater than z for a one-tailed test, since the percentage found in the survey (47.3%) is less than the claimed 50%.
To find the associated p-value, we look up the z-score in a standard normal distribution table or use a calculator with normal distribution capabilities. The p-value corresponds to the area to the right of a z of 2.98. Since normal distribution tables generally provide the area to the left, we subtract this value from 1 to get the area to the right. Using a standard normal distribution table or calculator, we find that the area to the left of z = 2.98 is approximately 0.9986. Therefore, the p-value = 1 - 0.9986 = 0.0014.