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 whether there has been a significant increase over the average 2010 figure of $18
reported in the neighborhood newspaper. State the hypotheses, alpha, df, and critical value.
Write a report including all necessary information about your results.

Answer 1

Answer on Question #49016 – 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.

Calculate whether there has been a significant increase over the average 2010 figure of $18 reported in the neighborhood newspaper. State the hypotheses, alpha, df, and critical value.

Solution

Hypotheses are

$H_{o}$: There is not a significant increase over the average 2010 figure of $18 reported in the neighborhood newspaper.

$H_{a}$: There is a significant increase over the average 2010 figure of $18 reported in the neighborhood newspaper.

**Evaluate**

$$\bar{x} = \frac{\sum (xf)}{n} = \frac{7 \cdot 4.5 + 7 \cdot 14.5 + 12 \cdot 24.5 + 5 \cdot 34.5 + 2 \cdot 44.5 + 1 \cdot 54.5}{34} = 21.85.$$ $$\sum (x)^2 = \sum (x^2 f) = 7 \cdot 4.5^2 + 7 \cdot 14.5^2 + 12 \cdot 24.5^2 + 5 \cdot 34.5^2 + 2 \cdot 44.5^2 + 1 \cdot 54.5^2 = 21698.46.$$ $$s = \sqrt{\frac{\sum (x)^2 - n\bar{x}^2}{n - 1}} = \sqrt{\frac{21698.46 - 34 \cdot (21.85)^2}{34 - 1}} = 12.87.$$

Hypotheses:

$H_{o}$: $\mu \leq 18$

$H_{a}$: $\mu > 18$

Decision Rule:

$\alpha = 0.05$

Degrees of freedom $n - 1 = 34 - 1 = 33$.

Critical t-score from t-table $t^* = 1.692$.

Reject $H_0$ if $t > 1.692$.

**Test Statistic:**

$$t = \frac {\bar {x} - \mu}{\frac {s}{\sqrt {n}}} = \frac {21.85 - 18}{\frac {12.87}{\sqrt {34}}} = 1.744.$$

**Decision (in terms of the hypotheses):**

Since $t = 1.744 > t^* = 1.692$ we reject $H_0$.

**Conclusion (in terms of the problem):**

There is a significant increase at 0.05 significance level over the average 2010 figure of $18 reported in the neighborhood newspaper.

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