Two friends, Karen and Jodi, work different shifts for the same ambulance service. They wonder if the different shifts average different numbers of calls. Looking at past records, Karen determines from a random sample of 40
shifts that she had a mean of 3.8
calls per shift. She knows that the population standard deviation for her shift is 1.2
calls. Jodi calculates from a random sample of 31
shifts that her mean was 4.6
calls per shift. She knows that the population standard deviation for her shift is 1.4
calls. Test the claim that there is a difference between the mean numbers of calls for the two shifts at the 0.05
level of significance. Let Karen's shifts be Population 1 and let Jodi's shifts be Population 2.
Step 2 of 3 :  Compute the value of the test statistic. Round your answer to two decimal places.

We are tasked with testing the claim that there is a difference between the mean number of calls for Karen's and Jodi's shifts at the 0.05 level of significance.

### Given dataWe are tasked with testing the claim that there is a difference between the mean number of calls for Karen's and Jodi's shifts at the 0.05 level of significance.

### Given 

- Karen's shifts (Population 1):
  - Sample size (\( n_1 \)) = 40
  - Sample mean (\( \bar{x}_1 \)) = 3.8
  - Population standard deviation (\( \sigma_1 \)) = 1.2

- Jodi's shifts (Population 2):
  - Sample size (\( n_2 \)) = 31
  - Sample mean (\( \bar{x}_2 \)) = 4.6
  - Population standard deviation (\( \sigma_2 \)) = 1.4

### Step 2: Compute the value of the test statistic

This is a two-sample \( z \)-test since we know the population standard deviations for both samples. The formula for the \( z \)-test statistic in comparing two means is:

\[
z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}
\]

Now, let’s plug in the values:

\[
z = \frac{(3.8 - 4.6)}{\sqrt{\frac{1.2^2}{40} + \frac{1.4^2}{31}}}
\]

Let's calculate this step-by-step.The value of the test statistic \( z \) is \(-2.54\), rounded to two decimal places.

Would you like further clarification or details on this process?

Here are 5 related questions to expand your understanding:
1. How do we interpret the \( z \)-test statistic in hypothesis testing?
2. What is the next step after finding the test statistic?
3. What does the negative sign of the \( z \)-value indicate in this context?
4. How does knowing the population standard deviation affect the test used?
5. What assumptions are made when performing a \( z \)-test for two populations?

**Tip**: When comparing two means, ensure that the test is appropriate based on whether population standard deviations are known or unknown, as this dictates the use of a \( z \)-test or \( t \)-test.

Two friends, Karen and Jodi, work different shifts for the same ambulance service. They wonder if the different shifts average different numbers of calls. Looking at past records, Karen determines from a random sample of 40 shifts that she had a mean of 3.8 calls per shift. She knows that the population standard deviation for her shift is 1.2 calls. Jodi calculates from a random sample of 31 shifts that her mean was 4.6 calls per shift. She knows that the population standard deviation for her shift is 1.4 calls. Test the claim that there is a difference between the mean numbers of calls for the two shifts at the 0.05 level of significance. Let Karen's shifts be Population 1 and let Jodi's shifts be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.

Test the claim that there's a difference in the mean number of calls for two ambulance service shifts. Using a two-sample z-test at a 0.05 significance level, this problem involves comparing the means from Karen's and Jodi's shifts using population standard deviations. Ideal for high school statistics or college-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation