Question 1204650: 1. The nine items of a sample had the following values: 45, 47, 50, 52, 48, 47, 49, 53, 51. Does the mean of the nine items differ significantly from the assumed population mean 47.5? (a = 0.05)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! find the mean and standad deviation of the sample.
i get meqan = 49.1111 and standard deviation = 2.6914
standard error = standard deviation / sqrt(sample size) = 2.6914 / sqrt(9) = .8971
t-score = (x-m)/s = (49.1111 - 47.5) / .8971 = 1.7959.
area under normal distributuion curve to the right of that t-score with 8 degrees of freedom = .0551
two tailed critical p-factor on each end of 95% confidence interval = .025.
this is higher than that, indicating the resuls are not significant.
there is not enough information to say that the population mean is not what is claimed to be.
i solved this with my ti-84 plus calculator and confirmed the answer through the use of the online statistical calculator at https://www.statskingdom.com/130MeanT1.html
here are the results from that calculator.
the results indicae that the p-value is 10.22.
the p-value i got from my ti-84 plus was .0511.
the .0511 p-value i got was for the high end tail of the confidence interval.
that would be alpha/2.
alpha is therefore twice that = 10.22
the critical alpha is .05 on both tails.
that's .025 for each tail.
when comparing p-values, you would either compare .0511 to .025 or .1022 to .05.
either way, the test p-value is higher than the critical p-value by a fairly wide margin.
the results from the online calculator tell you that the 95% confidence interval is 47.0977 to 51.1245.
49.1111 is within tht interval, indicating the results are not significant at .05 two tailed significance level.
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