Question 1139092: Can someone help me? I think I got confused while rounding.
Statistics students believe that the mean score on a first statistics test is 65. The instructor thinks that the mean score is higher. She samples 10 statistics students and obtains the scores:
Grades: 96, 68.4, 83.2, 66.5, 88, 83.2, 85.5, 96, 88, 96.
Test grades are believed to be normally distributed.
Use a significance level of 5%.
State the mean of the sample:
State the standard error of the sample means:
State the test statistic:
State the p-value:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! since you don't have the standard devisation of the population, you would need to use the stadard deviation of the sample to calculate the standard error of the sample means.
i used excel to get the mean and standard deviation because it's easier than manually calculating it, although i did show how the manual calculation of the standard deviation would be done.
these are the results from excel.
you have a sample mean of 85.08 and a sample standard deviation of 10.56975.
the standard error of the sample means is equal to the standard deviation of the sample, in this case, divided by the square root of the sample size.
that would result in standard error = 10.56975 / sqrt(10) = 3.382846 truncated to 6 decimal digits.
because you don't have the population standard deviation, you are using the t-score rather than the z-score.
the t-score of the sample is based on the formula t = (x - m) / s
t is the t-score
x is the raw score which is the mean of your sample.
m is the mean which is the assumed population mean.
s is the standard error of the distribution of sample means.
t-score formula becomes:
t = (85.08 - 65) / 3.382846 = 5.935830 truncated to 6 decimal digits.
since we're looking to see if the sample scores aee higher than the assumed population mean, we have a one tailed distribution at a significance level of .05.
looking up in the t-score tables for the critical t-score with 9 degrees of freedom at the .05 significance level for a one tailed distribution, we see that the critical t-score is equal to a t-score of 1.833.
since the t-score of our sample is 5.935830, it exceeds the critical t-score by a wide margin.
we can therefore concluded that the results of the t-test are significant and we would reject the null hypothesis that the mean of the population is 65.
the fact that the results are significant means that it is not very likely that the sample mean can be explained by normal variation in sample means of the same size.
the critical p-value is equal to .05
the sample p-value is equal to .000109563 truncated to 9 decimal digits.
this sample p-value is way smaller than the critical p-value.
this makes sense since the sample t-score is way higher than the critical t-score.
i used the TI-84 Plus calculator to determine the p-value of the sample since the tables are not as easy to determine it from.
the t-table i used just shows the critical t-score and is not designed to give you the sample t-score, nor the sample p-value.
the t-table i used can be found at http://www.ttable.org/
any questions, write to dtheophilis@gmail.com
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