We conduct a two-tailed t-test of hypotheses H0: µ = 120 H1: µ =/= 120.
The sample mean was 118.375 and the sample standard deviation was 2.801785145.
This gives a t statistic of ( 118.375 - 120 ) / 2.801785145 / sqrt(16) or t = -2.319949483
Asking Google sheets for the probability we get this value or something more extreme, i.e. the p-value, we get p-value = t.dist.2t(abs(-2.319949483), 15) = 0.03485256763, less than 0.05, and so we reject H0, and conclude corrective measures need to be taken.
Hope this helps! Let me know if I can explain another method if needed. I can do TI-83/84 as well if preferred.
Hilka L.
Thank you so much. I’d like for you to explain another method 😊08/11/23