The manager at the YMCA believes members are staying longer at the Y due to what they feel was a very successful marketing plan. Studies show the previous mean time per visit was 36 minutes, with a standard deviation = 11 minutes. A simple random sample of n = 200 visits is selected, and the current sample mean = 36.8 minutes. Test the manager's claim at a =0.05 using the p-value.
(please provide complete detailed answer)
You must be having an identity crisis! You've posted six different questions with six different names.
To test the manager's claim, we will perform a hypothesis test. The null hypothesis (H0) is that the mean time per visit is equal to the previous mean time of 36 minutes, and the alternative hypothesis (Ha) is that the mean time per visit has increased and is now greater than 36 minutes.
H0: μ = 36
Ha: μ > 36
We will use the Z-test, which requires the population standard deviation (σ) to be known. In this case, the standard deviation is given as 11 minutes.
First, we need to calculate the test statistic (Z-score):
Z = (X̄ - μ) / (σ / √n)
Where X̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
Given data:
X̄ = 36.8 minutes
μ = 36 minutes
σ = 11 minutes
n = 200 visits
Calculate the Z-score:
Z = (36.8 - 36) / (11 / √200)
Z = 0.8 / (11 / 14.14)
Z = 0.8 / 3.112
Z = 0.256
Next, we need to calculate the p-value, which represents the probability of obtaining a sample mean as extreme as the one observed (or more extreme) under the null hypothesis.
Since the alternative hypothesis is one-tailed (greater than), we need to find the probability of getting a Z-score greater than 0.256 using a standard normal distribution table or a calculator.
Using a standard normal table or a calculator, we find that the probability of obtaining a Z-score greater than 0.256 is approximately 0.399.
Since the p-value (0.399) is greater than the significance level (0.05), we fail to reject the null hypothesis. There is not enough evidence to support the manager's claim that the mean time per visit has increased. The data does not provide sufficient evidence to conclude that the marketing plan was successful in increasing the time spent at the YMCA.
It's important to note that this conclusion is based on the given data and the assumption that the sample is representative of the entire population.