Question

Before you took this course, you probably heard many stories about Statistics courses.  Oftentimes parents of students have had bad experiences with Statistics courses and pass on their anxieties to their children.  To test whether actually taking Statistics decreases students' anxieties about Statistics, a Statistics instructor gave a test to rate student anxiety at the beginning and end of his course.  Anxiety levels were measured on a scale of 0-10.  Here are the data for 16 randomly chosen students from a class of 180 students:

Pre-course anxiety level 7 6 9 5 6 7 5 7 6 4 3 2 1 3 4 2
Post-course anxiety level 4 3 7 3 4 5 4 6 5 3 2 2 1 3 4 3
Difference (Pre – Post) 3 3 2 2 2 2 1 1 1 1 1 0 0 0 0 -1

Do the data indicate that anxiety levels about Statistics decreases after students take Statistics?

State an appropriate hypothesis.

Was your test one-tail upper tail, one-tail lower tail, or two-tail?  Explain why you chose that kind of test in this situation.

Determine the critical value(s) and the standard test statistic.  Explain the steps you used to find these values.

Draw the standard normal distribution curve labeling the values found in (c).

Test your hypothesis and then write your conclusion using good English grammar.

Answer 1

Result:

Do the data indicate that anxiety levels about Statistics decreases after students take Statistics?

Paired t test used.

State an appropriate hypothesis.

$Null \ hypothesis H_0 : \mu_d = 0 \ Alternative \ hypothesis: H_1 : \mu_d > 0$

upper tail test

Was your test one-tail upper tail, one-tail lower tail, or two-tail? Explain why you chose that kind of test in this situation.

We are testing whether actually taking Statistics decreases students' anxieties about Statistics.

Test is one-tail upper tail.

Determine the critical value(s) and the standard test statistic. Explain the steps you used to find these values.

n=16, df = n-1=15

critical t value at 0.05 level =1.753

test statistic

$t= \frac { \mu_d}{ s_d/ \sqrt n}$

$t= \frac { 1.1475}{ 1.1250/ \sqrt 16}$

= 3.922

Draw the standard normal distribution curve labeling the values found in (c).

Distribution Plot T, df 15 3.922 0.4 0.3 0.2 0.1 0.05 0.0 1.753

Test your hypothesis and then write your conclusion using good English grammar.

Since test statistic 3.922 > critical value 1.753, Ho is rejected.

We conclude that taking Statistics decreases students' anxieties about Statistics.

Paired t Test
Data
Hypothesized Mean Difference	0
Level of significance	0.05
Intermediate Calculations
Sample Size	16
DBar	1.1250
Degrees of Freedom	15
SD	1.1475
Standard Error	0.2869
t Test Statistic	3.9217
Upper-Tail Test
Upper Critical Value	1.7531
p-Value	0.0007
Reject the null hypothesis