We can test the significance of the claim using a one-sample t-test. The null and alternative hypotheses are as follows:
Null Hypothesis (H0): The average time for the ambulance service to reach its destination is 10 minutes.
Alternative Hypothesis (H1): The average time for the ambulance service to reach its destination is not 10 minutes.
Given that we have a sample size of 36 calls with a mean of 11 minutes and a variance of 16 minutes, we can calculate the t-statistic using the formula:
t = (x̄ - μ) / (s / √n)
Where:
x̄ = sample mean (11 minutes)
μ = population mean (10 minutes)
s = sample standard deviation (√16 = 4 minutes)
n = sample size (36 calls)
Plugging in the values:
t = (11 - 10) / (4 / √36)
t = 1 / (4 / 6)
t = 1.5
Next, we need to find the critical t-value at α = 0.05 level of significance with 35 degrees of freedom (n-1 = 36-1 = 35). Using a t-distribution table or calculator, the critical t-value is approximately ± 2.030.
Since the calculated t-value (1.5) does not exceed the critical t-value (± 2.030), we do not have enough evidence to reject the null hypothesis. This means that we do not have sufficient evidence to conclude that the average time for the ambulance service to reach its destination is not 10 minutes at the 0.05 significance level.
An ambulance service claims that it takes on the average less than 10 minutes to
reach its destination in emergency calls. A sample of 36 calls has a mean of 11 minutes and the variance of16 minutes. Test the significance at 0.05 level
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