Question

An ambulance agency claims that the mean length of service times is 10 minutes. Investigator suspects that this claim is wrong. She takes a random sample of 20 services and finds the mean length of service times is 15 minutes. Assume that the service time of the ambulance follows normal distribution. 

i) What is the probability that the mean length of service times is more than and equal to 10 minutes.
ii) Find the 95% confidence interval for the standard deviation of the mean length of service times.

16 Feb 2018
Answer :
Word Count : 475
To solve this, we will address both parts of the question step-by-step. ### Given Information: - The claimed mean length of service times (\(\mu_0\)) = 10 minutes - The sample mean (\(\bar{x}\)) = 15 minutes - Sample size (n) = 20 - We will assume normal distribution for the service times. ### i) Probability that the mean length of service times is greater than or equal to 10 minutes: To compute this, we need to perform a hypothesis test and calculate the probability based on the standard normal distribution (Z-distribution) or t-distribution, depending on the scenario. The null hypothesis (\(H_0\)) is that the mean length of service times is 10 minutes, i.e., \(\mu = 10\). We will use a one-sample t-test to calculate the probability because the sample size is small (n = 20), and we don't have information about the population standard deviation. Steps: 1. Calculate
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