Question

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?

Answer 1

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 deviation

n= sample size

$\overline{x}$ = Sample mean

For 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=0.9660

Decision : 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 .