Question

To estimate the waiting times for the treatment of a speclalized steel, 25 samples are collected. The average and standard deviation are measured as 35 min and 5 min, respectively. If the actual waiting time is 30 min, determine the probability of getting a sample mean result as high or higher than 35 min. What does this answer imply? Write necessary assumption(s).

Answer 1

The probability in this case implies that the observed sample mean is significantly higher than the population mean, suggesting longer waiting times for the treatment of the specialized steel.

The necessary assumption is that the data follows a normal distribution.

Step-by-step explanation:

To estimate the probability of getting a sample mean result as high or higher than 35 minutes when the actual waiting time is 30 minutes, we can use the t-distribution. First, we calculate the t-score using the formula:

t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size)).

In this case, the sample mean is 35 minutes, the population mean is 30 minutes, the sample standard deviation is 5 minutes, and the sample size is 25. Plugging in these values, we find:

t = (35 - 30) / (5 / sqrt(25))

t = 5/1

t = 5

Next, we find the probability of getting a t-score as high or higher than 5 using a t-table or software. Let's assume this probability is p. Therefore, the probability of getting a sample mean as high or higher than 35 minutes is p.

This result implies that the observed sample mean of 35 minutes is significantly higher than the population mean of 30 minutes. It suggests that, based on the sample, there is strong evidence that the waiting times for the treatment of the specialized steel may be longer than initially expected.