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Can someone help me with another statistics question?
17.The winning team’s scores in 7 high school basketball games were recorded. If the sample mean is 9.7 points and the sample standard deviation is 0.25 points, find the 98% confidence interval of the true mean.
A)(7.889, 14.58)C)(9.40, 10.00)
B)(11.203, 14.56)D)(8.374, 11.89)
I don’t know where to start.
Should I use the confidence interval for variance formula or the confidence interval for standard deviation formula or some thing completely different?
And thank you so much for your help my statistics class is driving me crazy.
The sample mean, xbar = 9.7
the sample sd, s = 0.25.
n = 7.
The std error = s/sqrt(n) = 0.25/sqrt(7).
Now using the t distribution, with degrees of freedom, 7-1 = 6., look to the 98% CI critical value.
t_critical = 3.143
The the CI would be xbar +/- std error*t_critical
= 9.7 +/- 0.3 = [9.4, 10.0]
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