A notable indicator of a baby's health is the weight gained in the first year of the baby's life. Assume that the population of all such weight gains for baby girls is approximately normally distributed. A study claimed that the mean of this population is 5.31kg . As a practicing pediatrician, you want to test this claim. So, you select a random sample of 15 baby girls and record the weight each gained in their first year.

Follow the steps below to construct a 90% confidence interval for the population mean of all the weight gains for baby girls in their first year. Then state whether the confidence interval you construct contradicts the study's claim. Sample mean 5.538 sample standard deviation 1.921

To construct a 90% confidence interval for the population mean weight gain of baby girls in their first year, we will use the **t-distribution** since the sample size is small (15) and we don't know the population standard deviation. Here's the step-by-step process:

### Step 1: Identify the relevant information

- Sample mean (\( \bar{x} \)) = 5.538 kg
- Sample standard deviation (\( s \)) = 1.921 kg
- Sample size (\( n \)) = 15
- Confidence level = 90%

### Step 2: Determine the degrees of freedom

The degrees of freedom (df) for the \( t \)-distribution is:

\[
df = n - 1 = 15 - 1 = 14
\]

### Step 3: Find the critical value of \( t \)

For a 90% confidence level and 14 degrees of freedom, we need to find the critical value (\( t_{\alpha/2} \)) from the t-distribution table. The value of \( t \) corresponding to a 90% confidence interval with \( df = 14 \) is approximately:

\[
t_{\alpha/2} \approx 1.761
\]

### Step 4: Compute the standard error of the mean

The standard error of the mean (SE) is given by:

\[
SE = \frac{s}{\sqrt{n}} = \frac{1.921}{\sqrt{15}} \approx 0.496
\]

### Step 5: Calculate the margin of error (ME)

The margin of error (ME) is given by:

\[
ME = t_{\alpha/2} \times SE = 1.761 \times 0.496 \approx 0.874
\]

### Step 6: Construct the confidence interval

The confidence interval is:

\[
\left( \bar{x} - ME, \bar{x} + ME \right) = \left( 5.538 - 0.874, 5.538 + 0.874 \right) = \left( 4.664, 6.412 \right)
\]

### Step 7: Interpret the results

The 90% confidence interval for the mean weight gain of baby girls in their first year is \( (4.664, 6.412) \) kg.

#### Does this contradict the study's claim?

- The study claimed that the population mean weight gain is 5.31 kg.
- Since 5.31 kg falls within the 90% confidence interval \( (4.664, 6.412) \), **the confidence interval does not contradict the study's claim**.

---

Do you want further details or have any questions?

Here are 5 related questions to expand this topic:

1. How would the confidence interval change if the sample size increased?
2. What is the difference between using the z-distribution and t-distribution in confidence intervals?
3. How can we interpret a 95% confidence interval instead of a 90% one?
4. What does it mean if the study's claim falls outside the confidence interval?
5. How is the standard error affected by a larger or smaller sample size?

**Tip**: Always check if you're dealing with a small sample size (n < 30) to use the t-distribution for more accurate confidence intervals.

A notable indicator of a baby's health is the weight gained in the first year of the baby's life. Assume that the population of all such weight gains for baby girls is approximately normally distributed. A study claimed that the mean of this population is 5.31kg. As a practicing pediatrician, you want to test this claim. So, you select a random sample of 15 baby girls and record the weight each gained in their first year. Follow the steps below to construct a 90% confidence interval for the population mean of all the weight gains for baby girls in their first year. Then state whether the confidence interval you construct contradicts the study's claim. Sample mean 5.538 sample standard deviation 1.921

Learn how to construct a 90% confidence interval for the mean weight gain of baby girls in their first year of life. This problem uses sample data (mean = 5.538 kg, standard deviation = 1.921 kg) and applies the t-distribution to determine if the population mean aligns with a study's claim of 5.31 kg.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation