Macmillan Learning Suppose Diana, an educational researcher at a local university, wants to test the impact of a new Spanish course that integrates cultural-immersion teaching techniques with standard teaching practices. She selects a simple random sample of 62 freshmen and divides them into 31 pairs, matched on IQ and high school GPA. She randomly selects one member of each pair to take the new course, while the other member in the pair takes the traditional course. Next, Diana records the course grade, tallied on a scale from 0 to 4, for all sample members at the end of the semester, and she computes the difference in grades between the members in each matched pair by subtracting the traditional course grade from the new course grade. She wants to determine if the new Spanish course improves or weakens student performance. She runs a matched-pairs t-test to test the null hypothesis, Ho: =0, against the alternative hypothesis, H, 0, where is the mean course grade difference for the student population. The sample statistics for Diana's test are summarized in the table. Variable Sample μ description mean traditional course grade new course grade Xtrad=3.45773 Xnew=3.32772 Sample standard deviation Strad = 2.30766 Snew = 2.21794 SEnew difference (new - traditional) x= -0.13001 s = 0.34221 Standard error estimate SEtrad = 0.41447 0.39835 SE = 0.06146 Although Diana does not know the standard deviation of the underlying population of course grade differences, she assumes that the population is normally distributed because the sample data are symmetric, single-peaked, and contain no outliers. Compute the 1-statistic for Diana's matched-pairs t-test. Provide your answer with precision to three decimal places. 1= Compute the P-value of the 1-statistic using software. You may find some software manuals helpful. Provide your answer precise to three decimal places. P= Select the accurate statement regarding Diana's test decision if she tests at a significance level of a = 0.05. O Diana should fail to reject the null hypothesis because the mean grade difference is too small. 00000 Diana should fail to reject the null hypothesis because the P-value is greater than 0.025. Diana should reject the null hypothesis because the P-value is less than the value a = 0.05. Diana should reject the null hypothesis because the P-value is greater than the value a = 0.05. O Diana should fail to reject the null hypothesis because the P-value is greater than the value a = 0.05.

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Macmillan Learning
Suppose Diana, an educational researcher at a local university, wants to test the impact of a new Spanish course that integrates
cultural-immersion teaching techniques with standard teaching practices. She selects a simple random sample of 62 freshmen
and divides them into 31 pairs, matched on IQ and high school GPA. She randomly selects one member of each pair to take the
new course, while the other member in the pair takes the traditional course.
Next, Diana records the course grade, tallied on a scale from 0 to 4, for all sample members at the end of the semester, and she
computes the difference in grades between the members in each matched pair by subtracting the traditional course grade from
the new course grade. She wants to determine if the new Spanish course improves or weakens student performance. She runs a
matched-pairs t-test to test the null hypothesis, Ho: =0, against the alternative hypothesis, H, 0, where is the mean
course grade difference for the student population.
The sample statistics for Diana's test are summarized in the table.
Variable
Sample
μ
description
mean
traditional course grade
new course grade
Xtrad=3.45773
Xnew=3.32772
Sample standard
deviation
Strad = 2.30766
Snew = 2.21794
SEnew
difference (new - traditional)
x= -0.13001
s = 0.34221
Standard error
estimate
SEtrad = 0.41447
0.39835
SE = 0.06146
Although Diana does not know the standard deviation of the underlying population of course grade differences, she assumes that
the population is normally distributed because the sample data are symmetric, single-peaked, and contain no outliers.
Compute the 1-statistic for Diana's matched-pairs t-test. Provide your answer with precision to three decimal places.
1=
Compute the P-value of the 1-statistic using software. You may find some software manuals helpful. Provide your answer
precise to three decimal places.
P=
Select the accurate statement regarding Diana's test decision if she tests at a significance level of a = 0.05.
O Diana should fail to reject the null hypothesis because the mean grade difference is too small.
00000
Diana should fail to reject the null hypothesis because the P-value is greater than 0.025.
Diana should reject the null hypothesis because the P-value is less than the value a = 0.05.
Diana should reject the null hypothesis because the P-value is greater than the value a = 0.05.
O Diana should fail to reject the null hypothesis because the P-value is greater than the value a = 0.05.
Transcribed Image Text:Macmillan Learning Suppose Diana, an educational researcher at a local university, wants to test the impact of a new Spanish course that integrates cultural-immersion teaching techniques with standard teaching practices. She selects a simple random sample of 62 freshmen and divides them into 31 pairs, matched on IQ and high school GPA. She randomly selects one member of each pair to take the new course, while the other member in the pair takes the traditional course. Next, Diana records the course grade, tallied on a scale from 0 to 4, for all sample members at the end of the semester, and she computes the difference in grades between the members in each matched pair by subtracting the traditional course grade from the new course grade. She wants to determine if the new Spanish course improves or weakens student performance. She runs a matched-pairs t-test to test the null hypothesis, Ho: =0, against the alternative hypothesis, H, 0, where is the mean course grade difference for the student population. The sample statistics for Diana's test are summarized in the table. Variable Sample μ description mean traditional course grade new course grade Xtrad=3.45773 Xnew=3.32772 Sample standard deviation Strad = 2.30766 Snew = 2.21794 SEnew difference (new - traditional) x= -0.13001 s = 0.34221 Standard error estimate SEtrad = 0.41447 0.39835 SE = 0.06146 Although Diana does not know the standard deviation of the underlying population of course grade differences, she assumes that the population is normally distributed because the sample data are symmetric, single-peaked, and contain no outliers. Compute the 1-statistic for Diana's matched-pairs t-test. Provide your answer with precision to three decimal places. 1= Compute the P-value of the 1-statistic using software. You may find some software manuals helpful. Provide your answer precise to three decimal places. P= Select the accurate statement regarding Diana's test decision if she tests at a significance level of a = 0.05. O Diana should fail to reject the null hypothesis because the mean grade difference is too small. 00000 Diana should fail to reject the null hypothesis because the P-value is greater than 0.025. Diana should reject the null hypothesis because the P-value is less than the value a = 0.05. Diana should reject the null hypothesis because the P-value is greater than the value a = 0.05. O Diana should fail to reject the null hypothesis because the P-value is greater than the value a = 0.05.
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