Respuesta :
Answer:
[tex]t=\frac{(5216-5063)-0}{\sqrt{\frac{8131^2}{35}+\frac{4774^2}{35}}}}=0.096[/tex]
The p value for this case would be given by:
[tex]p_v =P(t_{68}>0.096)=0.92[/tex]
Since the p value is higher than the significance level we don't have enough evidence to conclude that mean for 4 year college is higher than the mean for the 2 year college
Step-by-step explanation:
Data given
[tex]\bar X_{1}=5063[/tex] represent the mean for the two year college
[tex]\bar X_{2}=5216[/tex] represent the mean for the 4 year college
[tex]s_{1}=4774[/tex] represent the sample standard deviation for the two year college
[tex]s_{2}=8131[/tex] represent the sample standard deviation for the 4 year college
[tex]n_{1}=35[/tex] sample size for the group 1
[tex]n_{2}=35[/tex] sample size for the group 2
[tex]\alpha=0.05[/tex] Significance level provided
t would represent the statistic
System of hypothesis
We want to verify if average enrollment at four-year colleges is higher than at two-year colleges in the United States, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{1}-\mu_{2}\leq 0[/tex]
Alternative hypothesis:[tex]\mu_{2} - \mu_{1}> 0[/tex]
Since we don't know the population deviations the statisitc is given by:
[tex]t=\frac{(\bar X_{2}-\bar X_{1})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)
And the degrees of freedom are given by [tex]df=n_1 +n_2 -2=35+35-2=68[/tex]
Replacing the info given we got:
[tex]t=\frac{(5216-5063)-0}{\sqrt{\frac{8131^2}{35}+\frac{4774^2}{35}}}}=0.096[/tex]
P value
The p value for this case would be given by:
[tex]p_v =P(t_{68}>0.096)=0.92[/tex]
Since the p value is higher than the significance level we don't have enough evidence to conclude that mean for 4 year college is higher than the mean for the 2 year college
Using the t-distribution, it is found that since the test statistic is less than the critical value, there is not enough evidence to conclude that average enrollment at four-year colleges is higher than at two-year colleges in the United States.
At the null hypothesis, it is tested if the average is not higher, that is, the result of the subtraction if not greater than 0.
[tex]H_0: \mu_F - \mu_T \leq 0[/tex]
At the alternative hypothesis, it is tested if it is higher, that is:
[tex]H_1: \mu_F - \mu_T > 0[/tex]
The standard errors for both samples are:
[tex]s_F = \frac{8131}{\sqrt{35}} = 1374.39[/tex]
[tex]s_T = \frac{4774}{\sqrt{35}} = 806.95[/tex]
The distribution of the differences has mean and standard deviation given by:
[tex]\overline{X} = \mu_F - \mu_T = 5216 - 5063 = 153[/tex]
[tex]s = \sqrt{s_F^2 + s_T^2} = \sqrt{1374.39^2 + 806.95^2} = 1593.77[/tex]
The test statistic is:
[tex]t = \frac{\overline{X} - \mu}{s}[/tex]
In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.
Hence, the value is:
[tex]t = \frac{153 - 0}{1593.77}[/tex]
[tex]t = 0.1[/tex]
The critical value, for a right-tailed test, as we are testing if the mean is greater than a value, with a significance level of 0.05 and 35 + 35 - 2 = 68 df, is [tex]t^{\ast} = 1.67[/tex]
Since the test statistic is less than the critical value, there is not enough evidence to conclude that average enrollment at four-year colleges is higher than at two-year colleges in the United States.
A similar problem is given at https://brainly.com/question/16402495
