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Question 1
A group of investors wants to develop a chain of fast-food restaurants. In determining potential costs
for each facility, they must consider, among other expenses, the average monthly electric bill. They
decide to sample some fast-food restaurants currently operating to estimate the monthly cost of
electricity. They want to be 90% confident of their results and want the error of the interval estimate to
be no more than $100. They estimate that such bills range from $600 to $2,500. How large a sample
should they take?
|According to a rule of thumb, σ = (1/4)(Range) = (1/4)(2500 - 600) = 475 |
|Confidence Level % = 90 |
|z- score = 1.645 |
|Population SD, σ = 475 |
|Error, E = 100 |
|Sample Size, N = (z * σ / E)^2 = (1.645 * 475 / 100)^2 = 61.05 |
|N = 62 |
Question 2
Suppose a study reports that the average price for a gallon of self-serve regular unleaded gasoline is
$3.16. You believe that the figure is higher in your area of the country. You decide to test this claim
for your part of the United States by randomly calling gasoline stations. Your random survey of 25
stations produces the following prices (all in $). Assume gasoline prices for a region are normally
distributed. Do the data you obtained provide enough evidence to reject the claim? Use a 1% level of
significance.
3.27 3.29 3.16 3.20 3.37
3.20 3.23 3.19 3.20 3.24
3.16 3.07 3.27 3.09 3.35
3.15 3.23 3.14 3.05 3.35
3.21 3.14 3.14 3.07 3.10
|For the given data, x-bar = 3.173 and s = 0.069 |
|n = 25 |
|μ = 3.16 |
|s = 0.069 |
|x-bar = 3.173 |
|Hypotheses: |
|Ho: μ ≤ 3.16 |
|Ha: μ > 3.16 |
|Decision Rule: |
|α = 0.01 |
|Degrees of freedom = 25 - 1 = 24 |
|Critical t- score = 2.492159469 |
|Reject Ho if t > 2.492159469 |
|Test Statistic: |
|SE = s/√n = 0.069/√25 = 0.0138 |
|t = (x-bar - μ)/SE = (3.173 - 3.16)/0.0138 = 0.942028986 |
|p- value = 0.177782159 |
|Decision (in terms of the hypotheses): |
|Since 0.942028986 < 2.492159469 we fail to reject Ho |
|Conclusion (in terms of the problem): |
|There is no sufficient evidence to reject the claim. The mean price may not be greater than $3.16. |
Question 3
Where do CFOs get their money news? According to Robert Half International, 47% get their money
news from newspapers, 15% get it from communication/colleagues, 12% get it from television, 11%
from the Internet, 9% from magazines, 5% from radio, and 1% do not know. Suppose a researcher
wants to test these results. She randomly samples 67 CFOs and finds that 40 of them get their
money news from newspapers. Does the test show enough evidence to reject the findings of Robert
Half International? Use a = .05.
|n = 67 |
|p = 0.47 |
|p' = 40/67 = 0.597 |
|Hypotheses: |
|Ho: p = 0.47 |
|Ha: p ≠ 0.47 |
|Decision Rule: |
|α = 0.05 |
|Lower Critical z- score = -1.9600 |
|Upper Critical z- score = 1.9600 |
|Reject Ho if |z| > 1.9600 |
|Test Statistic: |
|SE = √{p (1 - p)/n} = √(0.47 * (1 - 0.47)/67) = 0.0610 |
|z = (p'- p)/SE = (0.597014925373134 - 0.47)/0.0609746705424572 = 2.0831 |
|p- value = 0.0372 |
|Decision (in terms of the hypotheses): |
|Since 2.0831 > 1.9600 we reject Ho and accept Ha |
|Conclusion (in terms of the problem): |
|There is sufficient evidence to reject the findings of Robert Half International. It appears that the proportion of CFOs who get their money news from the |
|newspapers is not 47%. |
Question 4
To answer this question, use the Data Analysis Tool pack in Excel and select “t-Test: Two-Sample
Assuming Equal Variances” from the list of available tools. Explain your answer (how did you decide if
men spend more) and include the output table. Some studies have shown that in the United States,
men spend more than women buying gifts and cards on Valentine’s Day. Suppose a researcher
wants to test this hypothesis by randomly sampling nine men and 10 women with comparable
demographic characteristics from various large cities across the United States to be in a study. Each
study participant is asked to keep a log beginning one month before Valentine’s Day and record all
purchases made for Valentine’s Day during that one-month period. The resulting data are shown
below. Use these data and a 1% level of significance to test to determine if, on average, men actually
do spend significantly more than women on Valentine’s Day. Assume that such spending is normally
distributed in the population and that the population variances are equal.
Men Women
107.48 125.98
143.61 45.53
90.19 56.35
125.53 80.62
70.79 46.37
83.00 44.34
129.63 75.21
154.22 68.48
93.80 85.84
126.11
|For the 9 men, x-bar = 110.917 and s = 28.792 |
|For the 10 women, x-bar = 75.463 and s = 30.530 |
|n1 = 9 |
|n2 = 10 |
|x1-bar = 110.916667 |
|x2-bar = 75.463 |
|s1 = 28.7915925 |
|s2 = 30.5296526 |
|Hypotheses: |
|Ha: μ1 ≤ μ2 |
|Ha: μ1 > μ2 |
|Decision Rule: |
|α = 0.05 |
|Degrees of freedom = 9 + 10 - 2 = 17 |
|Critical t- score = 1.73960672 |
|Reject Ho if t > 1.73960672 |
|Test Statistic: |
|Pooled SD, s = √[{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] = √(((9 - 1) * 28.7915925228182^2 + (10 - 1) * 30.5296526347746^2)/(9 + 10 - 2)) = |
|29.72440432 |
|SE = s * √{(1 /n1) + (1 /n2)} = 29.7244043229287 * √((1/9) + (1/10)) = 13.65742128 |
|t = (x1-bar -x2-bar)/SE = (110.916666666667 - 75.463)/13.6574212771363 = 2.595926855 |
|p- value = 0.00941903 |
|Decision (in terms of the hypotheses): |
|Since 2.59592686 > 1.739606716 we reject Ho and accept Ha |
|Conclusion (in terms of the problem): |
|There is sufficient evidence that men actually do spend significantly more than women on Valentine’s Day. |
| |
|Data Analysis Output: |
|t-Test: Two-Sample Assuming Equal Variances |
| |
| |
| |
| |
| |
| |
| |
| |
|Men |
|Women |
| |
|Mean |
|110.9166667 |
|75.463 |
| |
|Variance |
|828.9558 |
|932.05969 |
| |
|Observations |
|9 |
|10 |
| |
|Pooled Variance |
|883.5402124 |
| |
| |
|Hypothesized Mean Difference |
|0 |
| |
| |
|df |
|17 |
| |
| |
|t Stat |
|2.595926855 |
| |
| |
|P(T ................
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