Question

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. 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 1

Answer: The mean difference is between 799586.3 and 803257.9.

Step-by-step explanation: To estimate the mean difference for confidence interval:

Find the statistic sample:

• d = value of 6th - value of 13th;
• Sample mean of difference: mean = ∑d / n
• Sample standard deviation: s = ∑(d - mean)² / n - 1;

For the traffic count, mean = 1835.8 and s = 1382607.3

The confidence interval is 90%, so:

α = $\frac{1-0.9}{2}$

α = 0.05

The degrees of dreedom are:

df = n - 1

df = 10 - 1

df = 9

Using a t-ditribution table, the t-score for α = 0.05 and df = 9 is: t = 1.833.

Error will be:

E = $t.\frac{s}{\sqrt{n} }$

E = 1.833.($\frac{1382607.3}{\sqrt{10} }$)

E = 801422.1

The interval is: mean - E < μ < E + mean

1835.8 - 801422.1 < μ < 1835.8+801422.1

-799586.3 < μ < 803257.9

The estimate mean difference in trafic count between 6th and 13th using 90% level of confidence is between 799586.3 and 803257.9.