Here are summary statistics for randomly selected weights of newborn girls: [tex]n=246, \bar{x}=33.1 \, \text{hg}, s=6.6 \, \text{hg}[/tex]. Construct a confidence interval estimate of the mean. Use a [tex]99\%[/tex] confidence level. Are these results very different from the confidence interval [tex]31.2 \, \text{hg} \ \textless \ \mu \ \textless \ 35.4 \, \text{hg}[/tex] with only 13 sample values, [tex]\bar{x}=33.3 \, \text{hg}[/tex], and [tex]s=2.5 \, \text{hg}[/tex]?

What is the confidence interval for the population mean [tex]\mu[/tex]?
[tex]\square \, \text{hg} \ \textless \ \mu \ \textless \ \square \, \text{hg}[/tex] (Round to one decimal place as needed.)



Answer :

To construct a confidence interval estimate of the mean weight of newborn girls at a 99% confidence level, we can follow these steps:

1. Identify the given values:
- Sample size [tex]\( n = 246 \)[/tex]
- Sample mean [tex]\( \bar{x} = 33.1 \)[/tex] hg
- Sample standard deviation [tex]\( s = 6.6 \)[/tex] hg
- Confidence level [tex]\( 99\% \)[/tex]

2. Determine the t critical value:
To find the t critical value for the given confidence level and degrees of freedom (df = [tex]\( n - 1 = 245 \)[/tex]), we can use a t-distribution table or a statistical software. The t critical value for a 99% confidence level with 245 degrees of freedom is approximately [tex]\( 2.596 \)[/tex].

3. Calculate the margin of error:
The margin of error (ME) is given by the formula:
[tex]\[ ME = t_{\text{critical}} \times \frac{s}{\sqrt{n}} \][/tex]
Substituting the known values:
[tex]\[ ME = 2.596 \times \frac{6.6}{\sqrt{246}} \][/tex]
[tex]\[ ME \approx 2.596 \times 0.4207 \approx 1.092 \][/tex]

4. Determine the confidence interval:
The confidence interval (CI) is calculated as:
[tex]\[ CI = \bar{x} \pm ME \][/tex]
So, the lower bound is:
[tex]\[ 33.1 - 1.092 \approx 32.0 \, \text{hg} \][/tex]
And the upper bound is:
[tex]\[ 33.1 + 1.092 \approx 34.2 \, \text{hg} \][/tex]

Thus, the confidence interval for the population mean [tex]\( \mu \)[/tex] is:
[tex]\[ 32.0 \, \text{hg} < \mu < 34.2 \, \text{hg} \][/tex]

### Comparison with Previous Confidence Interval

The previous confidence interval provided was [tex]\( 31.2 \, \text{hg} < \mu < 35.4 \, \text{hg} \)[/tex], which was based on:
- Sample size [tex]\( n = 13 \)[/tex]
- Sample mean [tex]\( \bar{x} = 33.3 \)[/tex] hg
- Sample standard deviation [tex]\( s = 2.5 \)[/tex] hg

The new confidence interval [tex]\( 32.0 \, \text{hg} < \mu < 34.2 \, \text{hg} \)[/tex] is narrower than the previous one. This is expected as the new sample size is much larger ([tex]\( n = 246 \)[/tex] vs. [tex]\( n = 13 \)[/tex]), leading to a more precise estimate of the population mean. Additionally, the sample means are quite similar, with [tex]\( \bar{x} = 33.1 \, \text{hg} \)[/tex] versus [tex]\( \bar{x} = 33.3 \, \text{hg} \)[/tex].

In conclusion:
[tex]\[ 32.0 \, \text{hg} < \mu < 34.2 \, \text{hg} \][/tex]
is the 99% confidence interval for the population mean weight of newborn girls based on the given sample.