Mathematics, 23.06.2020 18:01, 24chrim

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?

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answered: kittykat5105

(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.

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 $\mu_1$ = average enrollment at four-year colleges in the United States.

$\mu_2$ = average enrollment at two-year colleges in the United States.

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

Alternate Hypothesis, H_A : $\mu_1 \mu_2$ {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. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ t__n_1_+_n_2_-_2

where, $\bar X_1$ = average enrollment at four-year colleges = 5216

$\bar X_2$ = average enrollment at two-year colleges = 5069

s_1 = sample standard deviation at four-year colleges = 8141

s_2 = sample standard deviation at two-year colleges = 4773

n_1 = sample of four-year colleges surveyed = 35

n_2 = sample of two-year colleges surveyed = 35

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

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

= 0.092

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

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

answered: Guest

29. because you will divide each number of months by 12 to find the number of years. which would mean, 5, 12.5, 16, 45, 50, 72.5. which is crazy. anyhow, i think you find the median in the middle. and yea hope your test goes good

answered: Guest

step-by-step explanation:

6.48/4 4 goes into 6.48 1.62 times