Question

The British Department of Transportation studied to see if people avoid driving on Friday the 13^th.   They did a traffic count on a Friday and then again on a Friday the 13^th at the same two locations ("Friday the 13th," 2013). The data for each location on the two different dates is in following table:

Table: 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

Let ?₁= mean traffic count on Friday the 6th. Let ?₂ = mean traffic count on Friday the 13th. Estimate the mean difference in traffic count between the 6^th and the 13^th using a 90% level.

(i) Determine the sample mean of differences x¯di  Determine the sample mean of differences  x¯d {"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mstyle mathsize="14px"><mrow><mfenced><mi mathvariant="bold">i</mi></mfenced><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">Determine</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">the</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">sample</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">mean</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">of</mi><mo mathvariant="bold">&#xA0;</mo><mi mathvariant="bold">differences</mi><mo mathvariant="bold">&#xA0;</mo><mo mathvariant="bold">&#xA0;</mo><msub><mover><mi mathvariant="bold">x</mi><mo mathvariant="bold">&#xAF;</mo></mover><mi mathvariant="bold">d</mi></msub></mrow></mstyle></math>"}

Enter in decimal form to nearest tenth.  

(ii) Determine sample standard deviation of differences s_d

Enter in decimal form to nearest hundredth. Examples of correctly entered answers:

0.01    0.20     0.35      3.00

(iii) Enter the level of significance ? used for this test:

Enter in decimal form. Examples of correctly entered answers: 0.01    0.02    0.05    0.10

(iv) Determine degrees of freedom for the sample of differences df_d:

Enter value in integer form. Examples of correctly entered answers:

2    5    9    23    77

(v) Determine t - score associated with critical value: t_c

Enter in decimal form to nearest thousandth. Examples of correctly entered answers:

0.0011    0.020     0.500     0.371 2.000

(vi) Determine "error bound of the mean of difference" E

Enter value in decimal form rounded to nearest tenth. Examples of correctly entered answers:

2.0      0.3     1.6       11.7

(vii) Determine confidence interval of the mean difference ?_d

Enter lower bound value to nearest tenth, followed by < , followed by "?d" for mean difference, followed by <, followed by upper bound value to nearest tenth. No spaces between any characters. Use "negative" sign if necessary. Do not use italics or enter units of measure. Examples of correctly entered answers:

0.74<?d<0.78

13.14<?d<13.96

-9.72<?d<-8.08

(viii) Using the confidence interval, select the correct description of the result of the survey:

A. We estimate with 90% confidence that the true population mean traffic count between Friday the 6^th and Friday the 13th is between 1154.1 and 2517.5.

B. We estimate with 90% confidence that the true mean difference in traffic counts between Friday the 6^th and Friday the 13th falls outside 1154.1 and 2517.5.

C. We estimate with 90% confidence that the true sample mean traffic counts between Friday the 6^th and Friday the 13th is between 1154.1 and 2517.5.

D. We estimate with 90% confidence that the true mean difference in traffic counts between Friday the 6^th and Friday the 13th is between 1154.1 and 2517.5.

Enter letter corresponding to most correct answer

Answer 1

Using SPSS,

The difference in the mean may be estimated and tested at 90% level using a paired t test, since the observations on the 2 days are depenedent ( Since obtained from the same location ).

Conducting a paired t teast using these data yields the following results:

The mean difference estimated at 90% level has the value

(i) The sample mean of difference is the mean of the difference in the paired value in each sample observation = 1835.8

(ii) The sample standard deviation of differences sd = 1176.01

(iii) Level of significance ? used for this test = 0.10

(iv) Degrees of freedom for the sample of differences (dfd) = No. of paired observations - 1

= 10 - 1 = 9

(v)  t - score associated with critical value (tc)

= t = Sample mean of difference / [Its standard deviation / sqrt ( No. of paired observations)]

t = 4.936

(vi) The estimated mean difference Error bound = Confidence interval for actual mean difference

Substituting the lower and upper bounds, we obtain the error bound as 681.7

(vii) Confidence interval of the mean difference ?d :

1154.1 < < 2517.5

(viii) The correct description would be

D. We estimate with 90% confidence that the true mean difference in traffic counts between Friday the 6th and Friday the 13th is between 1154.1 and 2517.5.