Respuesta :
Answer:
[tex]t=\frac{2.59-2.34}{\sqrt{\frac{0.482^2}{25}+\frac{0.456^2}{30}}}}=1.963[/tex]
[tex]df=n_{A}+n_{B}-2=30+25-2=53[/tex]
[tex]p_v =2*P(t_{(53)}>1.963)=0.0549[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the two means are different at 10% of significance.
Step-by-step explanation:
Data given and notation
[tex]\bar X_{A}=2.34[/tex] represent the mean for the sample of night students
[tex]\bar X_{B}=2.59[/tex] represent the mean for the sample of day students
[tex]s_{A}=0.456[/tex] represent the sample standard deviation for the sample of night students
[tex]s_{B}=0.482[/tex] represent the sample standard deviation for the sample of day students
[tex]n_{A}=30[/tex] sample size selected for the night students
[tex]n_{B}=25[/tex] sample size selected for the day students
[tex]\alpha=0.1[/tex] represent the significance level for the hypothesis test.
t 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 two means are equal or no, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{A} = \mu_{B}[/tex]
Alternative hypothesis:[tex]\mu_{A} \neq \mu_{B}[/tex]
If we analyze the size for the samples both are <= than 30 and we don't know the population deviations, for this case is better apply a t test to compare means, and the statistic is given by:
[tex]t=\frac{\bar X_{B}-\bar X_{A}}{\sqrt{\frac{s^2_{B}}{n_{B}}+\frac{s^2_{A}}{n_{A}}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{2.59-2.34}{\sqrt{\frac{0.482^2}{25}+\frac{0.456^2}{30}}}}=1.963[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n_{A}+n_{B}-2=30+25-2=53[/tex]
Since is a two sided test the p value would be:
[tex]p_v =2*P(t_{(53)}>1.963)=0.0549[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the two means are different at 10% of significance.
Using the t-distribution, it is found that since the p-value of the test is of 0.0553 < 0.1, it can be concluded that the mean GPA of night students is different from the mean GPA of day students.
At the null hypothesis, we test their means are the same, that is:
[tex]H_0: \mu_{D} - \mu_{N} = 0[/tex]
At the alternative hypothesis, we test if the means are different, that is:
[tex]H_0: \mu_{D} - \mu_{N} \neq 0[/tex]
The standard errors are:
[tex]s_{D} = \frac{0.482}{\sqrt{25}} = 0.0964[/tex]
[tex]s_{N} = \frac{0.456}{\sqrt{30}} = 0.0833[/tex]
The distribution of the differences has:
[tex]\overline{x} = \mu_{D} - \mu_{N} = 2.59 - 2.34 = 0.25[/tex]
[tex]s = \sqrt{s_{D}^2 + s_{N}^2} = \sqrt{0.0964^2 + 0.0833^2} = 0.1274[/tex]
We have the standard deviation for the samples, hence, the t-distribution is used. The test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
In which [tex]\mu[/tex] is the value tested at the null hypothesis, for this problem [tex]\mu = 0[/tex], hence:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
[tex]t = \frac{0.25 - 0}{0.1274}[/tex]
[tex]t = 1.96[/tex]
Now, we have to find the p-value for the test, which is found using a two-tailed test(test if two values are different), with t = 1.96 and 25 + 30 - 2 = 53 df.
Using a t-distribution calculator, this p-value is of 0.0553.
Since the p-value of the test is of 0.0553 < 0.1, it can be concluded that the mean GPA of night students is different from the mean GPA of day students.
A similar problem is given at https://brainly.com/question/16736504
