Final answer:
To compare the salaries between Company A and B, we can use a hypothesis test, likely a two-sample t-test, relying on the known standard deviation and sample means. Calculating the 90th percentile for individual and average managers' salaries employs z-scores and standard errors from the normal distribution. Lastly, the median is a better central tendency measure than the mean in the presence of outliers, like in the case of one extremely high salary among many lower ones.
Step-by-step explanation:
Considering the information provided, we need to conduct a hypothesis test to determine whether the difference in sample means between Company A and Company B is statistically significant or not. We can undertake a two-sample t-test assuming that the standard deviation is the same for both companies since they are in the same industry. The known standard deviation for all mid-level professionals is $11,000, and we have the sample means for both companies. Using these values, the test will compare the sample means, and if the resulting p-value is less than the chosen significance level (typically 0.05), we would conclude that there is a significant difference in the pay of mid-level professionals between the two companies.
If you're interested in calculating percentiles for a salary distribution of managers in a restaurant chain, you will use the z-score formula for a normal distribution to find the z-score corresponding to the 90th percentile. Then, you would multiply this z-score by the standard deviation and add the result to the mean to get the 90th percentile for an individual manager's salary. To find this percentile for the average salary of the managers, you would use the standard error of the mean which is the standard deviation divided by the square root of the sample size. This accounts for the sample size in calculating the percentile for a sample mean.
In the context of the question involving a high income of $5,000,000 and numerous incomes of $30,000, it's crucial to recognize that the median is more robust to outliers than the mean and provides a better representation of the central tendency of the data when outliers are present.