Fight times for airplanes flying from San Francisco to Washington Dulles are kno
ID: 3253092 • Letter: F
Question
Fight times for airplanes flying from San Francisco to Washington Dulles are known to be Normally distributed with unknown mean and know standing sigma = 7 minutes. Listed below are airborne times (in minutes) for 10 randomly selected fights from San Francisco Washington Dulles airport: 270 272 273 290 274 275 266 275 271 285 a. Compute and interpret a 90% confidence interval for the mean airborne time for flights from San Francisco to Washington Dulles. Interpret the confident* level associated with the interval you computed. Airline officials insist that the mean time for this flight is mu = 270 minutes. What is your reaction to this claim? Use your result* from parts (a) and (b) in your answer. Perform a test of significance against the claim that mu = 270. State an appropriate alternative hypothesis (You may use your data for this step, but know that in practice, this is considered "fishing for significance'). Report the P - value of your test and form a conclusion you would give to airline officials.Explanation / Answer
Solution:-
Given,
count = 10
standard deviation = 7
Mean = 275.1
a).
90% confidence interval for mean airborne time for flights -
If (n < 30),
CI = x ± t/2 × (/n)
Where,
x = Mean
= Standard Deviation
= 1 - (Confidence Level/100)
t/2 = t-table value
CI = Confidence Interval
using the formula, CI = x ± t/2 × (/n)
= 275.1 ± 1.833 * (7/sqrt(10))
= 271.049 < µ < 279.151
b).
We are 90% confident that the mean airborne time for flight from San Fansisco to Washington Dulles, is between 271.049 to 279.151.
c).
As the claim by the airline official is outside the confidence interval that we have computed, so as per our calculation we tend to disagree with the claim.
d).
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 270
Alternative hypothesis: 270
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.10 (90% confidence interval). The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n) = 7 / sqrt(10) = 2.21359
DF = n - 1 = 10 - 1 = 9
t = (x - ) / SE = (275.1 - 270)/2.21359 = 2.30395 or 2.30
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the t statistic having 49 degrees of freedom is less than -2.30 or greater than 2.30.
We use the t Distribution Calculator to find P(t < -2.30) and P(t > 2.30)
The P-Value is 0.046999.
The result is significant at p < 0.10.
Interpret results. Since the P-value (0.046999) is smaller than the significance level (0.10), we can reject the null hypothesis.
Conclusion. Reject the null hypothesis. That is we have sufficient evidence to prove that the mean time is not equal to 270.
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