A group of 73 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic drinks they have in a typical week. The purpose of this study was to compare the drinking habits of the students at the college to the drinking habits of college students in general. In particular, the dean of students, who initiated this study, would like to check whether the mean number of alcoholic drinks that students at his college in a typical week differs from the mean of U.S. college students in general, which is estimated to be 4.73.
The group of 73 students in the study reported an average of 4.23 drinks per with a standard deviation of 3.75 drinks.
Find the p-value for the hypothesis test.
The p-value should be rounded to 4-decimal places.

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  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.

Learn more about p-value here:

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