Question

I surveyed each house on the block to determine how much we are spending on Halloween
treats. The distribution below represents the results in dollars.
$Spent f
50-59 1
40-49 2
30-39 5
20-29 12
10-19 7
0-9 7
Calculate the 90% confidence interval of the mean and present it in a sentence that shows
your understanding of the meaning of the confidence interval.

Answer 1

Answer on Question #49015 – Math – Statistics and Probability

I surveyed each house on the block to determine how much we are spending on Halloween treats. The distribution below represents the results in dollars.

$Spent f

50-59 1

40-49 2

30-39 5

20-29 12

10-19 7

0-9 7

Calculate the 90% confidence interval of the mean and present it in a sentence that shows your understanding of the meaning of the confidence interval.

Solution

The mean is

$$\mu = \frac {54.5 \cdot 1 + 44.5 \cdot 2 + 34.5 \cdot 5 + 24.5 \cdot 12 + 14.5 \cdot 7 + 4.5 \cdot 7}{1 + 2 + 5 + 12 + 7 + 7} = \frac {743}{34} = \$21.85.$$

Standard deviation is

$$\sigma = \sqrt {\frac {\sum f x ^ {2} - n \mu^ {2}}{n}}.$$ $$\sum f x ^ {2} = 54.5 ^ {2} \cdot 1 + 44.5 ^ {2} \cdot 2 + 34.5 ^ {2} \cdot 5 + 24.5 ^ {2} \cdot 12 + 14.5 ^ {2} \cdot 7 + 4.5 ^ {2} \cdot 7 = 21698.5.$$ $$\sigma = \sqrt {\frac {21698.5 - 34 \cdot 21.85 ^ {2}}{34}} = \$12.68.$$

The 90% confidence interval of the mean is

$$\begin{array}{l} \left(\mu - z _ {\frac {1 - \alpha}{2}} \frac {\sigma}{\sqrt {n}}; \mu + z _ {\frac {1 - \alpha}{2}} \frac {\sigma}{\sqrt {n}}\right) = \left(21.85 - z _ {0.05} \frac {12.68}{\sqrt {34}}; 21.85 + z _ {0.05} \frac {12.68}{\sqrt {34}}\right) \\ = \left(21.85 - 1.645 \cdot \frac {12.68}{\sqrt {34}}; 21.85 + 1.645 \cdot \frac {12.68}{\sqrt {34}}\right) = (\$18.27; \$25.43). \end{array}$$

With 90% confidence level, the mean spent of the block lies in the interval (\$18.27; \$25.43).

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