Question

Its claimed that the mean tenure for the chief of police was at least 9 years. A survey of metropolitan police departments found a sample tenure of x ¯=7.27 years for the chief of police position with a standard deviation of s=6.38 years.

If 85 were included in the sample what is the p-value for the hypothesis test?

Group of answer choices

between 0.005 and 0.01

between 0.01 and 0.25

between 0.025 and 0.05

between 0.05 and 0.10

greater than 0.10

Answer 1

Solution:

Test is lower tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 9

Xbar = 7.27

S = 6.38

n = 85

t = (7.27 - 9)/[6.38/sqrt(85)]

t = -2.50

df = n - 1 = 84

P-value = 0.0072

(by using t-table)

P-value is between 0.005 and 0.01.

Answer 2

Given:

• Claimed mean (μ₀): At least 9 years (so the null hypothesis is $H_0:\mu \ge 9$).

• Sample mean ($\bar{x}$): 7.27 years.

• Sample standard deviation (s): 6.38 years.

• Sample size (n): 85.

• 
  

Hypotheses:

• Null hypothesis ($H_0$): $\mu \ge 9$.

• Alternative hypothesis ($H_1$): $\mu <9$ (since we're testing if the mean is less than the claimed value).

Test Statistic (t-score):

Since the population standard deviation is unknown and the sample size is large ($n=85$), we use the t-test:

$$t=\frac{\bar{x}-{\mu }_0}{s\mathrm{/}\sqrt{n}}=\frac{7.27-9}{6.38\mathrm{/}\sqrt{85}}=\frac{-1.73}{0.692}\approx -2.50$$

Degrees of Freedom:

$$df=n-1=85-1=84$$

P-value:

This is a left-tailed test (because $H_1:\mu <9$). The p-value is the probability of observing a t-value less than $-2.50$ with 84 degrees of freedom.

From t-distribution tables or a calculator:

• For $t=-2.50$ and $df=84$, the p-value is between 0.005 and 0.01.

• 
  

Answer:

The p-value is between 0.005 and 0.01.