Question
Please, give the formula and steps to do this problem. Thank you
The average length of a flight by regional airlines in the United States has been reported as 299 miles. If a simple random sample of 30 flights by regional airlines If The average length of a flight by regional airlines in the United States has been reported as 299 miles. If a simple random sample of 30 flights by regional airlines were to have ȭ = 413/6 miles and s= 42.8 miles, would this tend to cast doubt on the reported average of 299 miles? Use a two-tail test and the 0.05 level of significance in arriving at your answer
The average length of a flight by regional airlines in the United States has been reported as 299 miles. If a simple random sample of 30 flights by regional airlines If The average length of a flight by regional airlines in the United States has been reported as 299 miles. If a simple random sample of 30 flights by regional airlines were to have ȭ = 413/6 miles and s= 42.8 miles, would this tend to cast doubt on the reported average of 299 miles? Use a two-tail test and the 0.05 level of significance in arriving at your answer
Answers
Is the random sample mean 41.5 (41 3/6) miles or 68.833 (413 divided by 6)? Either way, you do not have to do any calculation, since either value is over 5 standard deviations away from 299 miles.
However, to do a calculation, follow these steps.
Z = (mean1 - mean2)/ Standard Error (SE) of difference between means
SE of difference = Sq root of (SE1^2 + SE2^2)
SE = Standard Deviation/ Sq root of n-1
When only one measure of variability is available, it is used as the best estimate of SE of difference.
I hope this helps. Thanks for asking.
However, to do a calculation, follow these steps.
Z = (mean1 - mean2)/ Standard Error (SE) of difference between means
SE of difference = Sq root of (SE1^2 + SE2^2)
SE = Standard Deviation/ Sq root of n-1
When only one measure of variability is available, it is used as the best estimate of SE of difference.
I hope this helps. Thanks for asking.
WHAT IS WRONG WITH THESE STATEMENT?
The United Nations Organization declared that the world’s six-billionth inhabitant was born on October 12, 1999 in Bosnia.
The United Nations Organization declared that the world’s six-billionth inhabitant was born on October 12, 1999 in Bosnia.
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