Question

Is there any way to show this on Excel? I have the idea of how the problem is working but cannot figure out the calculation for z and p values.

Part I - Do Students Really Cheat? (30%) In a recent poll, 400 students were asked about their experiences with witnessing academic dishonesty among their classmates. Suppose 172 students admitted to witnessing academic dishonesty, 205 stated they did not and 23 had no opinion. Use the sign test and a significance of 0.05 to determine whether there is a difference between the number of students that have witnessed academic dishonesty compared to those that have not.

Answer 1

Ho: There is no difference between the number of students that have witnessed academic dishonesty compared to those that have not

Ha: There is a difference between the number of students that have witnessed academic dishonesty compared to those that have not

X+ = 172, X- = 205

Since n = X+ + X- = 172 + 205 = 377 > 25, we use normal approximation

α = 0.05

Critical z- score = 1.96

X = Min(172, 205) = 172

The test statistic z = [X + 0.5 – (n/2)]/[(√n)/2]

z = [172 + 0.5 – (377/2)]/[(√377)/2] = -1.648

Since |-1.648| < 1.96, we fail to reject Ho

Conclusion: There is no sufficient evidence of a difference between the number of students that have witnessed academic dishonesty compared to those that have not.

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