Question

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 5066 with a standard deviation of 4774. Of the 35 four-year colleges surveyed, the average enrollment was 5416 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.)

Part (d)State the distribution to use for the test. (Enter your answer in the form z or t_df where df is the degrees of freedom. Round your answer to two decimal places.)

• Part (e)

  What is the test statistic? (Round your answer to two decimal places.)
  ---Select--- z or t =

• Part (f)

  What is the p-value? (Round your answer to four decimal places.)

Answer 1

Answer:

Given,

Ho : u1 - u2 = 0

Ha : u1 - u2 < 0

consider,

Sp = sqrt(((n1-1)s1^2 + (n2-1)s2^2) / (n1+n2-2))

substitute values

= sqrt(((35-1)4774^2 + (35-1)8141^2) / (35+35-2))

= 6673.341

test statistic t = (x1 - x2)/Sp*sqrt(1/n1 + 1/n2)

substitute values

= (5066 - 5416)/6673.341*sqrt(1/35 + 1/35)

t = - 0.22

alpha = 0.05

degree of freedom = n1 + n2 - 2

= 35 + 35 - 2

= 68

P value = 0.413265

= 0.4133

Here we observe p value > alpha , so we fail to reject Ho.

So there is no sufficient evidence to support the claim.