12. A mocmoher wished to feat the effect of the addition of extra cafcium to yogurt on the "tastiness" of yogurt. Wab-hwo adulf wohanters were randomly divided into two groups of 31 subjects cach. Group 1 tasted yogurt cowaming the evin calcium, Gimup 2 tasted yogurt from the same batch as group I but without the added cakwam. Both groups rated the flavor on a scale of 1 to 10 , with I being "very unpleasant" and 10 being "very plearamt. The mean naring for group I was $$ \overrightarrow{1 j}=6.5 $$ with a standard deviation of s, $$ =1.5 $$. The mean rating for group 2 was $$ \overrightarrow{\mathrm{n}}=7.0 $$ with a standard deviation of $$ s, 2.0 $$. Let $$ \mu $$ and $$ \mu \mathrm{th} $$ represent the mean ratings we would observe for the entire population represented by the volunteers if all members of this population tasted, respectively, the yogurt with and without the added calcium. If we had used the more accurate software appratimation for the degrecs of frecdom, how would the $$ 90 \% $$ confidence interval compare to the one ognstructed with the more conservative value for degrees of freedom? It would be wider. Bevause $$ f $$ confidence intervals are constructed with sample standard deviations, we don't Know whether it would be wider, narrower, or the same. Cs. It would be narrower? Whether it would be wider, narrower, or stay the same depends on the sample sizes. If would not change.

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UpStudy AI · Accepted Answer

To analyze the problem, we need to understand how the degrees of freedom affect the confidence interval for the difference in means between the two groups.

### Known Conditions:
1. Group 1 (with added calcium):
   - Sample size $$ n_1 = 31 $$
   - Mean rating $$ \overline{x_1} = 6.5 $$
   - Standard deviation $$ s_1 = 1.5 $$

2. Group 2 (without added calcium):
   - Sample size $$ n_2 = 31 $$
   - Mean rating $$ \overline{x_2} = 7.0 $$
   - Standard deviation $$ s_2 = 2.0 $$

### Confidence Interval for the Difference in Means
The confidence interval for the difference in means $$ \mu_1 - \mu_2 $$ can be calculated using the formula:

$$
\text{CI} = (\overline{x_1} - \overline{x_2}) \pm t^* \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}
$$

Where:
- $$ t^* $$ is the critical value from the t-distribution based on the degrees of freedom.
- The degrees of freedom can be calculated using the formula for two independent samples:

$$
df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}}
$$

### Comparison of Confidence Intervals
1. **Using Conservative Degrees of Freedom**: This typically means using the smaller of $$ n_1 - 1 $$ and $$ n_2 - 1 $$ for the t-distribution, which can lead to a wider confidence interval because it uses a more conservative estimate of variability.

2. **Using More Accurate Degrees of Freedom**: The more accurate calculation of degrees of freedom (as shown above) may yield a higher value, which would result in a smaller critical value $$ t^* $$ for the same confidence level. This would lead to a narrower confidence interval.

### Conclusion
Based on the analysis, the correct answer to the question is:

**It would be narrower.**

This is because using a more accurate calculation for degrees of freedom typically results in a smaller critical value, leading to a narrower confidence interval.
Using a more accurate calculation for degrees of freedom results in a narrower 90% confidence interval.

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