Question

A group of 54 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 54 students in the study reported an average of 5.25 drinks per with a standard deviation of 3.98 drinks.Find the p -value for the hypothesis test. The p -value should be rounded to 4-decimal places.

Answer 1

Answer:

$t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601$  

P-value  

First we find the degrees of freedom given by:

$df = n-1= 54-1=53$

Since is a two-sided test the p value would be:  

$p_v =2*P(t_{53}>0.9601)=0.3414$  

Step-by-step explanation:

Data given and notation  

$\bar X=5.25$ represent the sample mean

$s=3.98$ represent the sample standard deviation

$n=54$ sample size  

$\mu_o =4.73$ represent the value that we want to test  

$\alpha$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean differs from 4.73, the system of hypothesis would be:  

Null hypothesis:$\mu =4.73$  

Alternative hypothesis:$\mu \neq 4.73$  

Since we know don't the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

$t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601$  

P-value  

First we find the degrees of freedom given by:

$df = n-1= 54-1=53$

Since is a two-sided test the p value would be:  

$p_v =2*P(t_{53}>0.9601)=0.3414$