It’s well established, we’ll assume, that lab rats require an average of 32 tria
ID: 3243038 • Letter: I
Question
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 = 34.89 and s = 3.02, 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. 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 = 34.89 and s = 3.02, 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. 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 = 34.89 and s = 3.02, 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.Explanation / Answer
(a) Given that X-Bar (the Mean of X) = 34.89 and s = 3.02, test the null hypothesis with t, using the .05 level of significance.
This is a 2-tailed t-test ....
Ho: mean = 32
Ha: mean 32
t-statistic = (34.89 - 32) / (3.02 / sqrt 7) = 2.532
df=7-1=6
P-value = p(t6<2.532)=0.0223
p=2*0.0223 =0.0466
Since the P-value < 0.05, REJECT the null hypothesis. this means that the shock traetment has caused change in the results water maze study
b) Construct a 95 percent confidence interval for the true number of trials required to learn the water maze
xbar=34.89
margin of error = 1.96
CI = 34.89 ± (1.96)(3.02)
CI = (28.97, 40.81)
(c) Interpret this confidence interval
we have 95 % confidenc e that the number of trials required is between 28.97 and 40.81
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