Question

/ The masses of 6 containers of Chow dog food are:


9.8

10.4

9.8

10

10.2

9.6












Find a 95% confidence interval for the mean mass of all such dog food. (Veronderstel ‘n benaderde normaalverdeling. / Assume an approximate normal distribution).

Answer 1

To calculate a 95% confidence interval for the mean mass of the dog food, we can follow these steps:

1. **Calculate the sample mean (\( \bar{x} \))**.
2. **Calculate the sample standard deviation (\( s \))**.
3. **Determine the appropriate t-score for 95% confidence**.
4. **Calculate the margin of error and the confidence interval**.

### Step 1: Calculate the Sample Mean

The sample mean can be calculated using the formula:

\[
\bar{x} = \frac{\sum x_i}{n}
\]

Where \( x_i \) are the sample observations and \( n \) is the sample size.

Given the masses of the containers:

9.8, 10.4, 9.8, 10.0, 10.2, 9.6

The sum of the masses is:

\[
9.8 + 10.4 + 9.8 + 10.0 + 10.2 + 9.6 = 60.8
\]

Now, calculate the mean:

\[
\bar{x} = \frac{60.8}{6} = 10.1333 \text{ (approximately)}
\]

### Step 2: Calculate the Sample Standard Deviation

The sample standard deviation \( s \) can be calculated using the formula:

\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}
\]

First, we calculate each deviation from the mean, square it, and then sum those squared deviations.

1. Deviations:

- \( 9.8 - 10.1333 \approx -0.3333 \)
- \( 10.4 - 10.1333 \approx 0.2667 \)
- \( 9.8 - 10.1333 \approx -0.3333 \)
- \( 10.0 - 10.1333 \approx -0.1333 \)
- \( 10.2 - 10.1333 \approx 0.0667 \)
- \( 9.6 - 10.1333 \approx -0.5333 \)

2. Squared deviations:

- \( (-0.3333)^2 \approx 0.1111 \)
- \( (0.2667)^2 \approx 0.0711 \)
- \( (-0.3333)^2 \approx 0.1111 \)
- \( (-0.1333)^2 \approx 0.0178 \)
- \( (0.0667)^2 \approx 0.0044 \)
- \( (-0.5333)^2 \approx 0.2844 \)

3. Sum of squared deviations:

\[
0.1111 + 0.0711 + 0.1111 + 0.0178 + 0.0044 + 0.2844 \approx 0.5999
\]

Now, plug into the standard deviation formula:

\[
s = \sqrt{\frac{0.5999}{6 - 1}} = \sqrt{\frac{0.5999}{5}} \approx \sqrt{0.11998} \approx 0.3463
\]

### Step 3: Determine the t-score

For a 95% confidence interval with \( n-1 = 5 \) degrees of freedom, we can find the t-score from the t-distribution table. The t-score for 95% confidence and 5 degrees of freedom is approximately:

\[
t_{0.025, 5} \approx 2.571
\]

### Step 4: Calculate the Margin of Error and Confidence Interval

The margin of error (ME) can be calculated using the formula:

\[
\text{ME} = t \cdot \frac{s}{\sqrt{n}}
\]

\[
\text{ME} = 2.571 \cdot \frac{0.3463}{\sqrt{6}} \approx 2.571 \cdot 0.1414 \approx 0.364
\]

Now, we can calculate the confidence interval:

\[
\text{CI} = \left( \bar{x} - \text{ME}, \bar{x} + \text{ME} \right)
\]

\[
\text{CI} = \left( 10.1333 - 0.364, 10.1333 + 0.364 \right) = \left( 9.7693, 10.4973 \right)
\]

So, the 95% confidence interval for the mean mass of all such dog food is approximately:

\[
(9.77, 10.50) \text{ (rounded to two decimal places)}
\]