1. ## auditory signal

The ability to recognize an auditory "signal" presented against a background of auditory "noise" is important to a particular job. A company tests for this ability by presenting 100 trials of auditory noise. During each trial the subject must state whether or not a signal has also been presented. A training program has been used to try to improve performance in this test. A random sample of 50 persons, who have had the training program, are administered the auditory test and they make an average of 32.2 errors.

a) The company's records indicated that in the past the mean number of errors has been 32.6 with a variance of 2.25. Does the training program improve test performance? (Use alpha less than or equal to 0.05)

b) With alpha less than or equal to 0.02, form a confidence interval to estimate the mean number of errors that would be made by all subjects who undergo the training program. [31.71 to 32.69]

2. Originally Posted by aptiva
The ability to recognize an auditory "signal" presented against a background of auditory "noise" is important to a particular job. A company tests for this ability by presenting 100 trials of auditory noise. During each trial the subject must state whether or not a signal has also been presented. A training program has been used to try to improve performance in this test. A random sample of 50 persons, who have had the training program, are administered the auditory test and they make an average of 32.2 errors.

a) The company's records indicated that in the past the mean number of errors has been 32.6 with a variance of 2.25. Does the training program improve test performance? (Use alpha less than or equal to 0.05)
The null hypothesis $H_0$ is that there is no change in the distribution of the
number of errors.

The critical z-score that a normal distributed RV exceeds with probability
0.95 is -1.64521. (a test with this critical value will reject the null hypothesis
if it were true 0.05 of the time, which is an alpha of 0.05)

The mean number of errors in a sample of size 50 persons under $H_0$ is 32.5,
and the SD of this is $\sqrt{2.25/50}\approx 0.212132$

Therefore we would reject the null hypothesis if the experiment had a mean
number of errors $< 32.6-0.212132*1.64521=32.251$

But the results gave a mean number of errors of 32.2 which is less than
32.251, so we rejsct the null hypothesis using this test.

RonL

3. Originally Posted by aptiva
The ability to recognize an auditory "signal" presented against a background of auditory "noise" is important to a particular job. A company tests for this ability by presenting 100 trials of auditory noise. During each trial the subject must state whether or not a signal has also been presented. A training program has been used to try to improve performance in this test. A random sample of 50 persons, who have had the training program, are administered the auditory test and they make an average of 32.2 errors.

b) With alpha less than or equal to 0.02, form a confidence interval to estimate the mean number of errors that would be made by all subjects who undergo the training program. [31.71 to 32.69]
For the given level of alpha the confidence interval in z-score will be
+/- 2.327. Therefore the interval will be

[32.2-2.327*0.2121.32.2+2.327*0.2121]=[31.71,32.69]

RonL