SOLUTION: Can you please help me solve this statistic problem? 6) It's well established, we'll assume, that lab rats require an average of 32 trials in a complex water maze before reachin

Question 769319: Can you please help me solve this statistic problem?
6) It's well established, we'll assume, that lab rats require an average of 32 trials in a complex water maze before reaching a learning criterion of three consecutive errorless trials. To determine whether a mildly adverse stimulus has any effect on performance, a sample of seven lab rats were given a mild electrical shock just before each trial.
(a) Given that X-Bar (the Mean of X) = 35.91 and s = 2.88, test the null hypothesis with t, using the .05 level of significance.
(b) Construct a 95 percent confidence interval for the true number of trials required to learn the water maze
(c) Interpret this confidence interval

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
It's well established, we'll assume, that lab rats require an average of 32 trials in a complex water maze before reaching a learning criterion of three consecutive errorless trials.
To determine whether a mildly adverse stimulus has any effect on performance, a sample of seven lab rats were given a mild electrical shock just before each trial.
--------------------
(a) Given that X-Bar (the Mean of X) = 35.91 and s = 2.88, test the null hypothesis with t, using the .05 level of significance.
---
Ho: u = 32
Ha: u # 32
-----
t(35.91) = (35.91-32)/[2.88/sqrt(7)] = 3.5920
p-value = 2*P(t > 3.5920 when df = 6) = 2*tcdf(3.5920,100,6) = 0.0115
---
Conclusion:: Since the p-value is less than 5%, reject Ho.
Cnclusion: The shock treatment has casused a change.
================================================================
(b) Construct a 95 percent confidence interval for the true number of trials required to learn the water maze
x-bar = 35.91
ME = 1.96 = 2.1335
-----
95% CI: 35.91-2.1335 < u < 35.91+2.1335
95% CI: 33.78 < u < 38.04
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(c) Interpret this confidence interval
We have 95% confidence that the number of trials required
is between 33.78 and 38.04
==============================
Cheers,
Stan H.

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