Question 185819This question is from textbook doane-Seward
: Please I am new to statistic and i need help with the questions below can you please help me. the questions is
A 20- minute consumer survey mailed to 500 adults aged 25-34 included a $5 Starbucks gift certificate. The same survey was mailed to 500 adults aged 25-34 without the gift certificate. There were 65 responses from the first group and 45 from the second group. (a) Perform a two-tailed test comparing the responses rates (proportions) at a=.05. (b) from a confidence interval for the difference of proportions, without pooling the samples. Does it include zero?
Are woman’s feet getting bigger? Retailers in the last 20 years have had to increase their stock of larger sizes. Wal-Mart Stores, Inc., and Payless Shoe Source, Inc., have been aggressive in stocking larger sizes, and Nordstrom’s reports that its larger sizes typically sell out first, assuming equal variances, at a=.025, do these random shoe size samples of 12 randomly chosen women in each age group show that women’s shoe sizes have increase? (See The Wall Street Journal, July 17, 2004.) Shoe Size
Born in 1980: 8 7.5 8.5 8.5 8 7.5 9.5 7.5 8 8 8.5 9
Born in 1960: 8.5 7.5 8 8 7.5 7.5 7.5 8 7 8 7 8
Block buster is testing a new policy of waiving all late fees on DVD rentals using a sample of 10 randomly chosen customers. (a) At a= .10, does the data show that the mean number of monthly rentals has increased? (b) is the decision close? (C Are you convinced?
Customer
Customer No Late Fee Late Fee
1 14 10
2 12 7
3 14 10
4 13 13
5 10 9
6 13 14
7 12 12
8 10 7
9 13 13
10 13 9
This question is from textbook doane-Seward
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A 20- minute consumer survey mailed to 500 adults aged 25-34 included a $5 Starbucks gift certificate. The same survey was mailed to 500 adults aged 25-34 without the gift certificate. There were 65 responses from the first group and 45 from the second group.
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(a) Perform a two-tailed test comparing the responses rates (proportions) at a=.05.
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Ho: p1-p2 = 0
Ha: p1-p2 is not 0
Using a 2=Proportion z-test on a TI-83 I get:
test statistic = 13.2116
p-value = 7.8x10^-40
Conclusion: since the p-value is less than 5%, reject Ho
The proportions are not equal
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(b) from a confidence interval for the difference of proportions, without pooling the samples. Does it include zero?
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Using a 2-Sample Z Interval on a TI-83 I get:
(0.4632 , 6568) which does not include zero
Conclusion: the proportions are not equal.
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Are woman’s feet getting bigger? Retailers in the last 20 years have had to increase their stock of larger sizes. Wal-Mart Stores, Inc., and Payless Shoe Source, Inc., have been aggressive in stocking larger sizes, and Nordstrom’s reports that its larger sizes typically sell out first, assuming equal variances, at a=.025, do these random shoe size samples of 12 randomly chosen women in each age group show that women’s shoe sizes have increase? (See The Wall Street Journal, July 17, 2004.) Shoe Size
Born in 1980: 8 7.5 8.5 8.5 8 7.5 9.5 7.5 8 8 8.5 9
Born in 1960: 8.5 7.5 8 8 7.5 7.5 7.5 8 7 8 7 8
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Sample mean for 1980: 8.2 ; Sample s = 0.6201
Sample mean for 1960: 7.7 ; sample s = 0.4502
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Ho: u(80)-u(60) = 0
Ha: u(80)-u(60) > 0
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Using a 2-Sample T Test on a TI-83 I get:
test statistic = 2.2605
p-value = 0.0175
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Conclusion: Since the p-value is less than 5%, reject Ho.
1980 sizes are larger than 1969 sizes
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Block buster is testing a new policy of waiving all late fees on DVD rentals using a sample of 10 randomly chosen customers. (a) At a= .10, does the data show that the mean number of monthly rentals has increased? (b) is the decision close? (C Are you convinced?
Customer
Customer No Late Fee Late Fee
1 14 10
2 12 7
3 14 10
4 13 13
5 10 9
6 13 14
7 12 12
8 10 7
9 13 13
10 13 9
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This problem is done the same way as the previous
Find the "no late fee" sample mean
Find the "late fee" sample mean
Ho: u(late)-u(not late) = 0
Ha: u(late)-u(not late) > 0
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Perform a 2-Sample T test to find:
the p-value
If it is less than 10% reject Ho.
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Cheers,
Stan H.
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