Normal or Parametric Data - slide 2

Try using a mu (mean) different from zero:

t.test(data.new[,2], mu= 11.066)
One Sample t-test
data:  data.new[, 2]
t = 1.2767, df = 41, p-value = 0.069
alternative hypothesis: true mean is not equal to 11.0695 percent confidence interval:
11.05714 11.07268
sample estimates:
mean of x
11.06491

This is much better. Now p > 0.05 meaning the data is not significantly different from 11.06.

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Transcript

Now, we could think of doing another computation, so, in this case, we're going to say instead of t-tests that we just did, we're going to now do another t-test, but we're going to say that the mu (the mean) is different from zero. And we're going to say that it's 11.066, which we saw was near the mean of what we found here. And now, whenever we do the one sided t-test, we see that we have a p-value of 0.069.  The alternative hypothesis is that the true mean is not equal to 11.066, etc.  We have a 95% confidence interval that it is this range. And here is again the mean value.  Now we see that this is a lot better because this p-value is much bigger than 0.05, that means the data is _not_ significantly different from 11.06, so 95% confidence interval means 0.5 for a p-value.  As our p-value is bigger than that, our probability of being in here is even better than 95%.