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
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.
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.
To determine whether the reported average of 299 miles is doubted based on the given sample mean and sample standard deviation, we can conduct a hypothesis test using a two-tailed test at the 0.05 level of significance.
The hypothesis statements for this test are as follows:
- Null hypothesis (H0): The true average length of regional airline flights is equal to 299 miles.
- Alternative hypothesis (Ha): The true average length of regional airline flights is not equal to 299 miles.
To conduct the hypothesis test, we need to calculate the test statistic, in this case, the t-score, and compare it to the critical value.
The formula to calculate the t-score is:
t = (ȭ - μ) / (s / sqrt(n))
where:
- ȭ is the sample mean (413/6 miles)
- μ is the hypothesized population mean (299 miles)
- s is the sample standard deviation (42.8 miles)
- n is the sample size (30 flights)
Substituting the given values into the formula:
t = ((413/6) - 299) / (42.8 / sqrt(30))
t ≈ (68.833 - 299) / (42.8 / sqrt(30))
t ≈ -230.167 / (42.8 / 5.48)
Calculating further:
t ≈ -230.167 / 7.842
t ≈ -29.36
Now, we need to compare the absolute value of the t-score to the critical value from the t-distribution table. Since this is a two-tailed test, we will divide the level of significance (0.05) by two and find the t-critical value for an alpha level of 0.025 (on each tail) with degrees of freedom of (n-1) = 29.
Using a t-distribution table or statistical software, the critical value for a t-test with 29 degrees of freedom and an alpha level of 0.025 is approximately 2.045.
Since the absolute value of the calculated t-score (-29.36) is greater than the critical value (2.045), we can reject the null hypothesis.
Therefore, the given sample data tends to cast doubt on the reported average of 299 miles, suggesting that the true average length of regional airline flights may not actually be 299 miles.