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# Stuck on how to figure this out. Please help on pointing me in the right direction.

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This is for Psy Statistics class and I'm struggling on how to solve problems like this one below?

Guest Jul 7, 2018

#1
+92805
+1

It has been a long time since I have done stats but I should be able to head you in the right direction.

First this site (that I found in our reference material sticky notes) is really useful

http://davidmlane.com/hyperstat/z_table.html

$$\sigma=0.5, \qquad \bar x=1.5\;seconds, \qquad n=25 \qquad \alpha=0.1\\~\\ H_o:\;\;\mu=1.8\\ H_A: \;\;\mu<1.8\\$$

NOTE:

On this graph below I would like to have added a second base lone where the mean is marked at 1.8 and the SD is 0.5.

If I was drawing by hand I would have done this.   But i used the webpage that i sited above :)

*        $$z_{crit}=-2.33$$

*         Decision Rule:     If  $$z_{test}<-2.33\qquad reject \quad H_o$$

$$\text{Calculate standard error}\\\sigma_{\bar x}=\frac{\sigma}{\sqrt n}=\frac{0.5}{\sqrt {25}}=\frac{0.5}{5}=0.1 \\ \qquad \text{Note: It is purely coincidental that this is the same as alpha}\\ Z_{test}=\frac{\bar X-\mu}{\sigma_{\bar x}}=\frac{1.5-1.8}{0.1}=\frac{-0.3}{0.1}=-3\\~\\ -3<-2.33 \;\;\;so\;\;\;H_0\;\; is\;\; rejected,\\ H_A \;\;is\;\;accepted$$

Conclusion: The hypothesis that the mean population is less than 1.8 is accepted (1% significance level)

This means that there is a less than 1% chance that the exercise program is NOT effective.

Conclusion: At a 1% significance level it can be concluded that the exercise program IS effective.

That should get you started I think.

Melody  Jul 7, 2018
#1
+92805
+1

It has been a long time since I have done stats but I should be able to head you in the right direction.

First this site (that I found in our reference material sticky notes) is really useful

http://davidmlane.com/hyperstat/z_table.html

$$\sigma=0.5, \qquad \bar x=1.5\;seconds, \qquad n=25 \qquad \alpha=0.1\\~\\ H_o:\;\;\mu=1.8\\ H_A: \;\;\mu<1.8\\$$

NOTE:

On this graph below I would like to have added a second base lone where the mean is marked at 1.8 and the SD is 0.5.

If I was drawing by hand I would have done this.   But i used the webpage that i sited above :)

*        $$z_{crit}=-2.33$$

*         Decision Rule:     If  $$z_{test}<-2.33\qquad reject \quad H_o$$

$$\text{Calculate standard error}\\\sigma_{\bar x}=\frac{\sigma}{\sqrt n}=\frac{0.5}{\sqrt {25}}=\frac{0.5}{5}=0.1 \\ \qquad \text{Note: It is purely coincidental that this is the same as alpha}\\ Z_{test}=\frac{\bar X-\mu}{\sigma_{\bar x}}=\frac{1.5-1.8}{0.1}=\frac{-0.3}{0.1}=-3\\~\\ -3<-2.33 \;\;\;so\;\;\;H_0\;\; is\;\; rejected,\\ H_A \;\;is\;\;accepted$$

Conclusion: The hypothesis that the mean population is less than 1.8 is accepted (1% significance level)

This means that there is a less than 1% chance that the exercise program is NOT effective.

Conclusion: At a 1% significance level it can be concluded that the exercise program IS effective.

That should get you started I think.

Melody  Jul 7, 2018
#2
+1

Thanks!!

Guest Jul 7, 2018