Compute test statistic (Independent t-score) Pooled variance S2p: Estimated standard error S(M1-M2) : Computed Independent t statistic: 2. Make a decision about the null hypothesis and state a conclusion. Decision: Are you going to reject the null or fail to reject the null Conclusion: ex. The independent variable did or did not have a significant effect Remember: if you reject, there was significance When people learn a new task, their performance usually improves when they are tested the next day, but only if they get 6 hours sleep (Sticckgold, et al., 2000). The following data demonstrate this phenomenon. The participants learned a visual discrimination task on one day. Half of the participants were allowed to have at least 6 hours of sleep and the other half were kept awake all night. 6 hours sleep No sleep n =14 n = 14 M = 72 M =65 SS = 932 SS = 706 Will the 6-hour sleep show a much higher performance rate? This time use a one-tail with alpha = .05 Show all of the steps. Remember use the attached t-distribution chart when obtaining your critical region.

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Description: Compute test statistic (Independent t-score) Pooled variance S2p: Estimated standard error S(M1-M2) : Computed Independent t statistic: 2. Make a decision about the null hypothesis and state a

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.3: Measures Of Spread

Problem 1GP

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Compute test statistic (Independent t-score)

Pooled variance S²_p:

Estimated standard error S(_M1-M2) :

Computed Independent t statistic:

2. Make a decision about the null hypothesis and state a conclusion.

Decision: Are you going to reject the null or fail to reject the null

Conclusion: ex. The independent variable did or did not have a significant effect

Remember: if you reject, there was significance

When people learn a new task, their performance usually improves when they are tested the next day, but only if they get 6 hours sleep (Sticckgold, et al., 2000). The following data demonstrate this phenomenon. The participants learned a visual discrimination task on one day. Half of the participants were allowed to have at least 6 hours of sleep and the other half were kept awake all night.

6 hours sleep No sleep

n =14 n = 14

M = 72 M =65

SS = 932 SS = 706

Will the 6-hour sleep show a much higher performance rate? This time use a one-tail with alpha = .05 Show all of the steps.

Remember use the attached t-distribution chart when obtaining your critical region.

Expert Solution

Step 1

The formula for pooled variance is given by :

$\begin{array}{rcl} {S^{2}}_{p} & = & \frac{(n_{1} - 1) . {s^{2}}_{1} + (n_{2} - 1) . {s^{2}}_{2}}{n_{1} + n_{2} - 2} \\ = & \frac{S S_{1} + S S_{2}}{n_{1} + n_{2} - 2} \end{array}$

Given the sum squares, the standard deviation is calculated by :

$s = \sqrt{\frac{S S}{n - 1}}$

$\begin{array}{rcl} s_{1} & = & \sqrt{\frac{932}{13}} \\ = & 8.467 \\ s_{2} & = & \sqrt{\frac{706}{13}} \\ = & 7.369 \end{array}$

$\begin{array}{rcl} {S^{2}}_{p} & = & \frac{932 + 706}{14 + 14 - 2} \\ = & \frac{1638}{26} \\ = & 63 \end{array}$

The estimated standard error is :

$\begin{array}{rcl} S . E (M_{1} - M_{2}) & = & \sqrt{\frac{{S^{2}}_{p}}{n_{1}} + \frac{{S^{2}}_{p}}{n_{2}}} \\ = & \sqrt{\frac{63 + 63}{14}} \\ = & \sqrt{9} \\ = & 3 \end{array}$

To compute the test statistics, first define the hypothesis.

$H_{o} : M_{1} = M_{2} H_{1} : M_{1} > M_{2}$

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