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In: Statistics and Probability

This assignment covers two-sample hypothesis tests. As with Calculator HW 5, our goal is to understand...

This assignment covers two-sample hypothesis tests. As with Calculator HW 5, our goal is to understand the five step method of hypothesis testing, as well as the calculator functions.     Your work must include:

1. Clear statement of hypotheses, with the correct parameter(s)
2. An indication of the test used
3. The test statistic and p-value
4. An indication of the statistical decision (i.e. whether or not to reject Ho)
     along with an explanation.
5. An interpretation of the statistical decision in the context of the problem.

           *   *   *   *   *   *   *   *   *   *   *   *   *   *   *   *   *   *   *   *   *   *

#1)   A sample of 12 in-state undergraduate school programs for State A has a mean tuition of $24,000 with a standard deviation of $4,600. In State B, a sample of 16 in-state undergraduate programs has a mean of $28,000 with a standard deviation of $5,200. Assume that in-state tuitions are normally distributed for both states. Using a .05 significance level, does this sample data provide evidence that the mean tuition are different for the two states?

Solutions

Expert Solution

We have to test for two population means. The data is normally distributed and the population SD are unknown. So we use t-test. Also since samples are from two different states, they are independent. We assume they have equal population variance. We will conduct an independent samples t-test.

Since we checking for difference not if one is small or not it is two tailed.

Step 1

Null: The mean tuitions fees is same for the two states.

Alternative: The mean tuitions fees is different for the two states.

Step 2

Independent samples t-test for difference of population means.

Pooled variance =

Test Stat =

p-value = 2P ( > T.S. ) ........using t-dist tables

Null :
Alternative:
State A (1) State B (2)
Mean 24000 28000
SD 4600 5200
Var 21160000 27040000
n 12 16
Pooled var 24552308
SE 1892.233
Step 3 Test Stat -2.1139
alpha 0.05
df 26
p-value 0.0443 P(t26>2.11)=0.0221
Since p-value < 0.05
Step 4 Decision Reject null hypothesis at 5%.
Step 5 Conclude There is significant evidence to conclude that the mean tuition fees is different for the two states.

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