A statistics teacher wants to assess whether her remedial tutoring has been effe

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.

Explanation / Answer

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)

=

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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