Question 1142140: The claim is that for a smartphone carrier's data speeds at airports, the mean is u=13.00 Mbps. The sample size is n=12 and the test statistic is t= - 1.999.
P-value=
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! degrees of freedom = sample size - 1 = 12 - 1 = 11.
t(-1.99) with 11 degrees of freedom has 3.59133847% of the area under the normal distribution curve to the left of it.
whether this is sufficient to refute the claim that the average data speed at airport is 13.00 megabits per second depends on what the cutoff is for statistical significance.
at .99 confidence level, with 11 degrees of freedom, the critical t-score would be -3.105806514.
at .95 confidence level, with 11 degrees of freedom, the critical t-score would be -2.200985143.
at .90 confidence level, with 11 degrees of freedom, the critical t-score would be -1.795884781.
the critical t-score is calculated based on a two tailed normal distribution.
what this means is that we find the low side critical t-score by taking half of the alpha.
for example, at .01 confidence level, the alpha is 1 - .99 = .01
the alpha to find the critical t-score is equal to .005, which is half the critical alpha.
this gives you half of the alpha on the low side of the confidence interval and half of the alpha on the high side of the confidence interval.
based on the test statistic, the claim that the average speed of the data at the airport cannot be refuted at .99 and .95 confidence level, but can be refuted at .90 confidence level.
this is because the t-score of the test is -1.99.
this t-score does not exceed the critical t-score at the .99 and.95 confidence level thresholds, but does exceed the critical t-score at the .90 confidence level threshold.
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