1. A group of teachers has developed a new reading program designed to help second graders learn to read. In order to test the effectiveness of their program, they check the scores on their students achievement tests and compare them to the nationwide average. The 20 students in their class had a mean score of 179. The achievement test has been used for quite a long time, and it has been established that the population mean for the test is 145 with a population standard deviation of 50.

• a) State the null hypothesis and the alternative hypothesis.
• b) Find the z-score
• c) Find the p-value
• d) Do the scores for the sample of 20 students provide enough evidence to conclude that the test is effective at a .01 level of significance?
• e) Professional educators consider that if a reading program would produce a difference of 30 points in mean scores for this test, it would represent a significant improvement in teaching reading. What is the power of this test to detect such an improvement at a .01 level of significance?

a)

then the null hypothesis is

and the alternative hypothesis is

Since the teachers want to establish that their reading program is superior to the standard reading programs, this will be a one tailed test

b) Since the achievement test has been given so often, we can be pretty sure that the 50 is a very good estimate of the population standard deviation. If we have the population standard deviation, we use z-scores. To find the z-score, use the formula

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where

and

This gives us

c) A z-score of 3.04 in Table A corresponds to an area to the left of the z-score of

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.9988

to get a p-value of

1 - .9988

= .0012

d) The probability of being this far into the tail end of the bell shaped curve when the null hypothesis is true is .0012 which is less than 1%. We reject the null hypothesis and conclude that the reading program represents a significant improvement at even a 1% level of significance.

e) To find the power of the test to determine a 30 point improvement in mean scores at a 1% level of significance, we first determine how large a mean will result in the rejection of the null hypothesis at a 1% level of significance. The critical value of z for a 1% level of significance in a one tailed test is found from Table A or Table D to be

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2.326

The next question is how large of a sample mean will result in this big of a t-score. If we solve the equation

for the sample mean, we get

= 145 + 2.326(11.18)

= 171

This is what the picture would look like if the null hypothesis were true

Now let us assume that the alternative hypothesis is true. If there is a difference of 30 points in the mean score of the achievement tests then that would represent a mean score on the achievement test for students from the reading program of 175. In this case the picture would actually look like

The power of the test is the probability that we will reject Ho under these conditions. To find that we compute the z-score using the formula

The closest z-score in Table A is -.36 which corresponds to an area of

.3594

This will be the probability that the sample mean would find itself in the region to the left of the critical point which would be the region of acceptance of the null hypothesis. The probability of rejecting the null hypothesis would then be

1 - .3594

= .6406

which says that there is about as 64% chance that this test would detect a difference of 30%.

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