Question 888535
sample mean = 6.9629
sample standard deviation = 1.9000
sample size = 35
degrees of freedom = 35 - 1 = 34
standard error = 1.9/sqrt(35) = .32116


assumed population mean is 8.
population standard deviation is not known.
use t-score rather than z-score.


t-score = (sample mean minus population mean) / standard error = (6.9629 - 8) / .32116 = -3.23


significance level of 2% for a one way confidence limit = an alpha of .02


critical t score for an alpha of .02 with 34 degrees of freedom would be plus or minus 2.13.


anything that exceeds that is statistically significant.


that means that a t score of less than -2.13 and a t score greater than +2.13 would be considered statistically significant and more then likely not due to chance variation between samples of size 35 .


the t score of the data is well beyond that and so the sample is considered statistically significant.


the probability of getting a t score less than -3.23 is  .00137 which is significantly less than the significance level alpha of .02.


you would reject the null hypothesis that the average hours of sleep for the population was 8 hours or more.


even if you made this into a 2 tailed significance level, the alpha would have been .01 and the critical t factor would have been plus or minus 2.44.
-3.23 is still well below -2.44.