Question

indy 500 qualifier speeds the speeds in miles per hour of six randomly selected qualifiers for the indianapolis 500 (in 2012) are listed below. estimate the mean qualifying speed with 95% confidence. round your answers to at least four decimal places.
222.891 222.929 223.422 223.684
224.037 225.172

Explanation:

Step1: Calculate sample mean

Sample speeds: 222.891, 222.929, 223.422, 223.684, 224.037, 225.172
Sum = 222.891 + 222.929 + 223.422 + 223.684 + 224.037 + 225.172 = 1342.135
Mean $\bar{x} = \frac{1342.135}{6} \approx 223.689167$

Step2: Calculate sample standard deviation

Deviations from mean:
222.891-223.689167≈-0.798167, 222.929-223.689167≈-0.760167, 223.422-223.689167≈-0.267167,
223.684-223.689167≈-0.005167, 224.037-223.689167≈0.347833, 225.172-223.689167≈1.482833
Squared deviations sum: $(-0.798167)^2 + (-0.760167)^2 + (-0.267167)^2 + (-0.005167)^2 + (0.347833)^2 + (1.482833)^2 ≈ 0.63707 + 0.57785 + 0.07138 + 0.00003 + 0.12099 + 2.1988 ≈ 3.60612$
Sample variance $s^2 = \frac{3.60612}{6-1} ≈ 0.721224$
Sample standard deviation $s = \sqrt{0.721224} ≈ 0.849249$

Step3: Find t-critical value

df = 6-1=5, 95% confidence → $t_{0.025,5}=2.5706$

Step4: Calculate margin of error

$E = t \times \frac{s}{\sqrt{n}} = 2.5706 \times \frac{0.849249}{\sqrt{6}} ≈ 2.5706 \times 0.3464 ≈ 0.8904$

Step5: Compute confidence interval

Lower: $223.689167 - 0.8904 ≈ 222.7988$
Upper: $223.689167 + 0.8904 ≈ 224.5796$

Answer:

The 95% confidence interval for the mean qualifying speed is (222.7988, 224.5796) mph.