Q:

# The Scottsdale fire department aims to respond to fire calls in 4 minutes or less, on average. Response times are normally distributed with a standard deviation of 1 minute. Would a sample of 18 fire calls with a mean response time of 4 minutes 30 seconds provide sufficient evidence to show that the goal is not being met at α = .01?

Accepted Solution

A:
Answer with explanation:Let $$\mu$$ be the population mean.As per given , we have$$H_0:\mu \leq4\\\\ H_a: \mu >4$$Since the alternative hypothesis is right-tailed , so the test is a right-tailed test.Also, population standard deviation is given $$\sigma=1$$ , so we perform one-tailed z-test.Test statistic : $$z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$$, where $$\mu$$ = Population mean$$\sigma$$ = Population standard deviationn= sample size$$\overline{x}$$ = Sample meanFor n= 18 , $$\overline{x}=4.50$$ , $$\sigma=1$$ , $$\mu =4$$, we have $$z=\dfrac{4.5-4}{\dfrac{1}{\sqrt{18}}}\approx2.12$$P-value (for right tailed test): P(z>2.12) = 1-P(z≤ 2.12)  [∵ P(Z>z)=1-P(Z≤z)]\=1- 0.0340=0.9660Decision : Since P-value(0.9660) > Significance level (0.01), it means we are failed to reject the null hypothesis.[We reject null hypothesis if p-value is larger than the significance level . ]Conclusion : We do not have sufficient evidence to show that the goal is not being met at α = .01 .