Solution :
a) degrees of freedom = n - 1 = 40 - 1 = 39
t/2,df = t0.01,39 = 2.426
Margin of error = E = t/2,df * (s /n)
= 2.426 * (5.7 / 40)
Margin of error = E = 2.19
The 98% confidence interval estimate of the population mean is,
± E
= 68.4 ± 2.19
= ( 66.21, 70.59 )
b) We are 98% confident that the true mean speed of all cars between 66.21 mi / hr and 70.59 mi / hr.
The recorded speeds (in mi/hr) is observed from a sample of 40 cars traveling on the...
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5. The recorded speeds (in mi/hr) is observed from a sample of 40 cars traveling on the 101 freeway near La Conchita. The sample has a mean of 68.4 mi/hr and a standard deviation of 5.7 mi/hr (based on data from Sigalert). You must do all of your calculations on the TI calculator. DO NOT Use the Stats -> Tests for any of the problems. a. Find a 98% confidence interval estimate for the mean speed of all cars. [5...
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