A student at a four-year college claims that average enrollment at four-year colleges is higher than at two-year colleges in the United States. Two surveys are conducted. Of the 35 two-year colleges surveyed, the average enrollment was 5069 with a standard deviation of 4773. Of the 35 four-year colleges surveyed, the average enrollment was 5216 with a standard deviation of 8141. Conduct a hypothesis test at the 5% level. NOTE: If you are using a Student's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

Required:
a. State the distribution to use for the test.
b. What is the test statistic?
c. What is the p-value?

Respuesta :

Answer:

a) Since we have sample data and we don't know the population deviation we can use the t distribution in order to test the hypothesis

t would represent the statistic

b) [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)  

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{(5216-5069)-0}{\sqrt{\frac{4773^2}{35}+\frac{8141^2}{35}}}}=0.0921[/tex]  

c) [tex]p_v =P(t_{68}>0.0921)=0.463[/tex]  

Since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean of enrollment is significantly higher for four year college than for two year college

Step-by-step explanation:

Information given

[tex]\bar X_{1}=5069[/tex] represent the mean for two year colleges

[tex]\bar X_{2}=5216[/tex] represent the mean for four year college

[tex]s_{1}=4773[/tex] represent the sample standard deviation for 1  

[tex]s_{2}=8141[/tex] represent the sample standard deviation for 2  

[tex]n_{1}=35[/tex] sample size for the group 2  

[tex]n_{2}=35[/tex] sample size for the group 2  

[tex]\alpha=0.05[/tex] Significance level provided

Part a

Since we have sample data and we don't know the population deviation we can use the t distribution in order to test the hypothesis

t would represent the statistic

Part b

We want to test if the mean of enrollment for four year college is higher than for two year college, the system of hypothesis would be:  

Null hypothesis:[tex]\mu_{2}-\mu_{1} \leq 0[/tex]  

Alternative hypothesis:[tex]\mu_{2} - \mu_{1} > 0[/tex]  

The statistic 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)  

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{(5216-5069)-0}{\sqrt{\frac{4773^2}{35}+\frac{8141^2}{35}}}}=0.0921[/tex]  

Part c

The p value would be given by:

[tex]p_v =P(t_{68}>0.0921)=0.463[/tex]  

Since the p value is higher than the significance level we don't have enough evidence to conclude that the true mean of enrollment is significantly higher for four year college than for two year college

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