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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) The test statistics that would be used here Two-sample t-test statistics distribution.

(b) The value of t-test statistic is 0.092.

(c) P-value of the test statistics is more than 40%.

Step-by-step explanation:

We are given that 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.

Let [tex]\mu_1[/tex] = average enrollment at four-year colleges in the United States.

[tex]\mu_2[/tex] = average enrollment at two-year colleges in the United States.

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu_1 \leq \mu_2[/tex]      {means that the average enrollment at four-year colleges is higher than at two-year colleges in the United States}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu_1 > \mu_2[/tex]      {means that the average enrollment at four-year colleges is higher than at two-year colleges in the United States}

(a) The test statistics that would be used here Two-sample t-test statistics distribution because we don't know about population standard deviation;

                              T.S. =  [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex]  ~ [tex]t__n_1_+_n_2_-_2[/tex]

where, [tex]\bar X_1[/tex] = average enrollment at four-year colleges = 5216

[tex]\bar X_2[/tex] = average enrollment at two-year colleges = 5069

[tex]s_1[/tex] = sample standard deviation at four-year colleges = 8141

[tex]s_2[/tex] = sample standard deviation at two-year colleges = 4773

[tex]n_1[/tex] = sample of four-year colleges surveyed = 35

[tex]n_2[/tex] =  sample of two-year colleges surveyed = 35

Also, [tex]s_p= \sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }[/tex]  =  [tex]\sqrt{\frac{(35-1)\times 8141^{2}+(35-1)\times 4773^{2} }{35+35-2} }[/tex] = 6672.98

So, the test statistics  =  [tex]\frac{(5216-5069)-(0)}{6672.98 \times \sqrt{\frac{1}{35}+\frac{1}{35} } }[/tex]  ~ [tex]t_6_8[/tex]

                                     =  0.092

(b) The value of t-test statistic is 0.092.

(c) P-value of the test statistics is given by the following formula;

                 P-value = P([tex]t_6_8[/tex] > 0.092) = More than 40% as this value is not reflected in the t-table.

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