# Solved (Free): It is known that the duration of consultation with patients in a psychiatry clinic is normally distributed with an average of 35 minutes

#### ByDr. Raju Chaudhari

Apr 16, 2021

It is known that the duration of consultation with patients in a psychiatry clinic is normally distributed with an average of 35 minutes and a standard deviation of 5 minutes. One of the doctors working in this clinic claimed that the duration of the consultation is not 35 minutes. For this purpose, the consultation duration of 10 clients was examined and found to be 42 minutes. Test this claim at 0.05 significance level.

#### Solution

Given that $n = 10$, sample mean $\overline{x} = 42$, the sample standard deviation is $\sigma = 5$.

#### Step 1 Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : \mu = 35$ against $H_1 : \mu \neq 35$ ($\text{two-tailed}$)

#### Step 2 Test Statistic

The test statistic for testing above hypothesis testing problem is

 \begin{aligned} Z& =\frac{\overline{x} -\mu}{\sigma/\sqrt{n}}. \end{aligned}
The test statistic $Z$ follows $N(0,1)$ distribution.

#### Step 3 Significance Level

The significance level is $\alpha = 0.05$.

#### Step 4 Critical Value(s)

As the alternative hypothesis is $\text{two-tailed}$, the critical value of $Z$ $\text{are}$ $-1.96 and 1.96$ (from Normal Statistical Table).

The rejection region (i.e. critical region) is $\text{Z < -1.96 or Z > 1.96}$.

#### Step 5 Computation

The test statistic under the null hypothesis is

 \begin{aligned} Z_{obs}&=\frac{ \overline{x} -\mu_0}{\sigma/\sqrt{n}}\\ &= \frac{42-35}{5/ \sqrt{10 }}\\ &= 4.427 \end{aligned}

#### Step 6 Decision

The test statistic is $Z_{obs} =4.427$ which falls $\text{inside}$ the critical region, we $\text{reject}$ the null hypothesis at $\alpha = 0.05$ level of significance.

#### Step 6 Decision

$p$-value Approach:

This is a $\text{two-tailed}$ test, so the p-value is the
area to the $\text{extreme}$ of the test statistic ($Z_{obs}=4.427$) is p-value = $0$.

The p-value is $0$ which is $\text{less than}$ the significance level of $\alpha = 0.05$, we $\text{reject}$ the null hypothesis at $\alpha =0.05$ level of significance.

There is sufficient evidence to conclude the duration of the consultation is not 35 minutes.