Question 698306: Please help me.
t distribution related problem.
Acme corporation manufactures light bulbs. The CEO claims that an average light bulb lasts 300 days. A researcher randomly select 15 bulbs for testing. The sampled bulbs last an average of 290 days, with a standard deviation of 50 days. If the CEO's claims were true, what is the probability that 15 randomly selected bulbs would have an average life of no more than 290 days?
Note: It require to compute the t score.
Answer by Positive_EV(69)   (Show Source):
You can put this solution on YOUR website! The t-statistic for this data is  , where x is the mean of the sample of bulbs, mu is the population mean, s is the standard deviation of the sample, and n is the number of items in the sample. Here, x = 290, mu = 300, s = 50, and n = 15, so:
 . So, the t-score is -.7746.
The t-distribution changes in shape depending on how many items are in the sample. You have to use the t-distribution corresponding to the appropriate number of degrees of freedom. For probability calculations, the number of degrees of freedom is n - 1, so here you need the t-distribution with 14 degrees of freedom.
You'll need either a table or a utility to calculate the probability that t < -.7746 with 14 degrees of freedom. I use http://stattrek.com/online-calculator/t-distribution.aspx as a calculator for this. This calculator returns the probability that, assuming the population mean is true, the t-value is less than the t-value obtained With 14 degrees of freedom and a t score of -.7746, the table returns a probability of the bulbs lasting less than 290 days on average of .2257 assuming the mean life of the bulbs is 300 days.
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