Question

select the correct answer from each drop-down menu.
alice wanted to compare birthday cake prices in two supermarkets. she wrote down the daily prices for a few days and found that the difference of the two sample means is $4.01. the standard deviation of the difference of the sample means is $5.46.
the 95% confidence interval of the sample mean difference is between and .
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Explanation:

Step1: Recall 95% CI formula

For a 95% confidence interval, we use \( \text{Mean} \pm 1.96 \times \text{Standard Deviation} \) (approximate for large samples, using the empirical rule or z - value for 95% confidence). Here, the mean difference (\(\bar{x}\)) is $4.01, and the standard deviation of the difference (LXI2) is $5.46.

Step2: Calculate the lower bound

Lower bound \(= \bar{x}-1.96\times s\)
Substitute \(\bar{x} = 4.01\) and \(s = 5.46\):
\(4.01-1.96\times5.46=4.01 - 10.7016=- 6.6916\approx - 6.69\)

Step3: Calculate the upper bound

Upper bound \(=\bar{x}+1.96\times s\)
Substitute \(\bar{x} = 4.01\) and \(s = 5.46\):
\(4.01 + 1.96\times5.46=4.01+10.7016 = 14.7116\approx14.71\)

Answer:

The 95% confidence interval of the sample mean difference is between \(-\$6.69\) and \(\$14.71\) (or more precise values based on exact calculation: lower bound \(\approx - 6.69\), upper bound \(\approx14.71\)). If we use more precise calculation: \(1.96\times5.46 = 10.7016\), so lower bound \(4.01-10.7016=-6.6916\), upper bound \(4.01 + 10.7016 = 14.7116\). Rounding to two decimal places, the interval is between \(-\$6.69\) and \(\$14.71\).