SOLUTION: The British Department of Transportation studied to see if people avoid driving on Friday the 13th. They did a traffic count on a Friday and then again on a Friday the 13th at th

Question 1187095: The British Department of Transportation studied to see if people avoid driving on Friday the 13th. They did a traffic count on a Friday and then again on a Friday the 13th at the same two locations ("Friday the 13th," 2013). The data for each location on the two different dates is in table #9.2.6. Estimate the mean difference in traffic count between the 6th and the 13th using a 90% level.
Table #9.2.6: Traffic Count
Dates 6th 13th
1990, July 139246 138548
1990, July 134012 132908
1991, September 137055 136018
1991, September 133732 131843
1991, December 123552 121641
1991, December 121139 118723
1992, March 128293 125532
1992, March 124631 120249
1992, November 124609 122770
1992, November 117584 117263

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to estimate the mean difference in traffic count with a 90% confidence level:
1. **Calculate the Differences:** Find the difference between the traffic count on the 6th and the 13th for each pair of dates. (6th - 13th)
* 1990, July: 139246 - 138548 = 698
* 1990, July: 134012 - 132908 = 1104
* 1991, September: 137055 - 136018 = 1037
* 1991, September: 133732 - 131843 = 1889
* 1991, December: 123552 - 121641 = 1911
* 1991, December: 121139 - 118723 = 2416
* 1992, March: 128293 - 125532 = 2761
* 1992, March: 124631 - 120249 = 4382
* 1992, November: 124609 - 122770 = 1839
* 1992, November: 117584 - 117263 = 321
2. **Calculate the Sample Mean Difference (d̄):** Sum the differences and divide by the number of pairs (n = 10).
d̄ = (698 + 1104 + 1037 + 1889 + 1911 + 2416 + 2761 + 4382 + 1839 + 321) / 10
d̄ = 18358 / 10
d̄ = 1835.8
3. **Calculate the Sample Standard Deviation of the Differences (sd):**
First, find the squared difference from the mean for each difference and sum them.
Then, divide the sum by (n-1) and take the square root.
sd ≈ 1220.17
4. **Find the t-score:** For a 90% confidence level and 9 degrees of freedom (n-1 = 10-1 = 9), the t-score (from a t-table or calculator) is approximately 1.833.
5. **Calculate the Margin of Error:**
Margin of Error = t-score * (sd / √n)
Margin of Error = 1.833 * (1220.17 / √10)
Margin of Error ≈ 706.85
6. **Calculate the Confidence Interval:**
Confidence Interval = d̄ ± Margin of Error
Confidence Interval = 1835.8 ± 706.85
Confidence Interval ≈ (1128.95, 2542.65)
**Conclusion:**
We are 90% confident that the true mean difference in traffic count between the 6th and the 13th is between approximately 1128.95 and 2542.65. Since the interval is positive, this suggests that traffic counts tend to be lower on Friday the 13th.

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