Respuesta :
We have large mu = 4.73, n = 73, x bar = 4.23, and s = 3.75. The test statistic is t = (x bar - large mu)/(s/n) = (4.23 - 4.73)/(3.75/73).
The p-value for t = -1.14, d. f = n -1 = 72, and a two-tailed test is performed with the Excel function TDIST(1.14, 72,
A statistical measure called a p-value is used to verify a hypothesis against observed data. If the null hypothesis is true, the probability of getting the observed results is what is measured by a p-value. The observed difference's statistical significance is greater when the p-value is lower.
What is meant by a p-value of 0.05?
The probability that the null hypothesis is correct is P > 0.05.The probability that the alternative hypothesis is correct is equal to 1 minus the P value. A test result that is statistically significant (P Â 0.05) indicates that the test hypothesis is false or ought to be rejected. A P value of greater than 0.05 indicates that there was no effect.
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