Respuesta :
Answer:
[tex]t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601[/tex]
P-value
First we find the degrees of freedom given by:
[tex] df = n-1= 54-1=53[/tex]
Since is a two-sided test the p value would be:
[tex]p_v =2*P(t_{53}>0.9601)=0.3414[/tex]
Step-by-step explanation:
Data given and notation
[tex]\bar X=5.25[/tex] represent the sample mean
[tex]s=3.98[/tex] represent the sample standard deviation
[tex]n=54[/tex] sample size
[tex]\mu_o =4.73[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] 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:[tex]\mu =4.73[/tex]
Alternative hypothesis:[tex]\mu \neq 4.73[/tex]
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:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (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:
[tex]t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601[/tex]
P-value
First we find the degrees of freedom given by:
[tex] df = n-1= 54-1=53[/tex]
Since is a two-sided test the p value would be:
[tex]p_v =2*P(t_{53}>0.9601)=0.3414[/tex]
Using the t-distribution, it is found that the p-value for this hypothesis test is of 0.0656.
At the null hypothesis, it is tested if the mean at this college is the same as the mean of U.S. college students in general, hence:
[tex]H_0: \mu = 4.73[/tex]
At the alternative hypothesis, it is tested if it is different, hence:
[tex]H_1: \mu \neq 4.73[/tex]
We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
The parameters are:
[tex]\overline{x}[/tex] is the sample mean.
[tex]\mu[/tex] is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.
For this problem, the values of the parameters are:
[tex]\overline{x} = 5.75, \mu = 4.73, n = 54, s = 3.98[/tex]
Hence, the value of the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{5.75 - 4.73}{\frac{3.98}{\sqrt{54}}}[/tex]
[tex]t = 1.88[/tex]
The p-value is found using a two-tailed test, as we are testing if the mean is different of a value, with t = 1.88 and 54 - 1 = 53 df.
Using a t-distribution calculator, this p-value is of 0.0656.
A similar problem is given at https://brainly.com/question/15463898
