Question

A statistics teacher wants to assess whether her remedial tutoring has been effective for her 16 students (n = 16). To evaluate the changes that occur during the year, students’ grades were measured at the beginning and at the end of the year. Their grades revealed an average improvement of MD = 8 points with a sum of squares for the difference scores of SSD = 930. Conduct the appropriate hypothesis test to determine if the data are sufficient to conclude that there was significant IMPROVEMENT? Use a one-tailed test with ? = 0.01.

What is the null hypothesis (H0)? A) µD = 0 B) µD ? 0 C) µD ? 0 D) µD > 0

What is the alternative hypothesis (H1)? A) µD = 0 B) µD ? 0 C) µD > 0 D) µD ? 0

What are the degrees of freedom (df)? A) df = 15 B) df = 16 C) df = 14 D) df = 17

What is the tcritical value? A) tcritical = +2.947 B) tcritical = +2.602 C) tcritical = -2.947 D) tcritical = -2.602

What is the variance of the difference scores (s2D)? A) s2D = 116.25 B) s2D = 58.13 C) s2D = 66.43 D) s2D = 62.00

What is the estimated standard error for the sample mean difference (sMD)? A) sMD = 1.91 B) sMD = 2.03 C) sMD = 1.97 D) sMD = 2.11

What is the t-statistic? A) t = 3.79 B) t = 4.06 C) t = 4.19 D) t = 3.94

What is the conclusion? (A) Fail to reject H0: Tutoring significantly improves grades B) Fail to reject H0: Tutoring DOES NOT significantly improve grades. C) Reject H0: Tutoring significantly improves grades. D) Reject H0: Tutoring DOES NOT significantly improve grades.

Answer 1

What is the null hypothesis

null hypothesis contains equality
hence

A) is correct

What is the alternative hypothesis

since we want to know if there was significant improvement

C) µD > 0

What are the degrees of freedom (df)?

df = n-1= 16-1 = 15

A) df = 15

critical value

one-tailed test with ? = 0.01.

= t.inv(0.99,15)

=

2.60248

B) tcritical = +2.602  

variance = SS_d/(n-1) = 930 /15 =62

standard error = sd/sqrt(n) = sqrt(62)/sqrt(16)
= 1.9685019685

C) sMD = 1.97

t-stat = (Xbar -mu)/se
= (8 - 0)/1.9685019685
= 4.0640

B) t = 4.06

What is the conclusion?

since TS > critical value

C) Reject H0: Tutoring significantly improves grades

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