Question

A survey of 100 similar-sized hospitals revealed a mean daily census in the pediatrics service of 27 with a standard deviation of 6.5. Researchers test whether these data provide sufficient evidence to indicate the mean is greater than 25. Use α=.05. Give 1. the hypotheses, 2. appropriate test, 3. decision rule, 4. calculated test statistic, and 5. conclusion with a comparison to the critical value or alpha.

Answer 1

Answer:

We conclude that the mean is greater than 25.  

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 25

Sample mean, $\bar{x}$ = 27

Sample size, n = 100

Alpha, α = 0.05

Sample standard deviation, s = 6.5

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 25\\H_A: \mu> 25$

We use One-tailed(right) z test to perform this hypothesis.

Formula:

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

Putting all the values, we have

$z_{stat} = \displaystyle\frac{27 - 25}{\frac{6.5}{\sqrt{100}} } = 3.07$

Now, $z_{critical} \text{ at 0.05 level of significance } = 1.645$

Since,  

$z_{stat} > z_{critical}$

We reject the null hypothesis and accept the alternate hypothesis.

Thus,  the mean is greater than 25.