Question

The department of public safety has an old memo stating that the number of accidents per week at a hazardous intersection varies according to a Normal distribution, with a mean of 2.2 and a standard deviation of 1.4. Department officials implemented a new safety plan, heavier police patrols and new signs, to see if they could reduce the average number of accidents at this intersection. They recorded the number of accidents per week for 52 weeks. They find that the average over that period was two accidents per week. What is the P ‑value for the test of H 0 : μ = 2.2 against H a : μ < 2.2 ?

Answer 1

Answer:

P-value = 0.1515

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 2.2

Sample mean, $\bar{x}$ = 2

Sample size, n = 52

Alpha, α = 0.05

Population standard deviation, σ = 1.4

First, we design the null and the alternate hypothesis

$H_{0}: \mu =2.2\\H_A: \mu < 2.2$

We use one-tailed(left) z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{2 - 2.2}{\frac{1.4}{\sqrt{52}} } = -1.03$

Now, we calculate the p-value from the normal standard table.

P-value = 0.1515