Question

A teacher instituted a new reading program at school. After 10 weeks in the​ program, it was found that the mean reading speed of a random sample of 20 second grade students was

94.4 wpm. What might you conclude based on this​ result? Select the correct choice below and fill in the answer boxes within your choice.

​(Type integers or decimals rounded to four decimal places as​ needed.)

A. A mean reading rate of 94.4 wpm is not unusual since the probability of obtaining a result of 94.4 wpm or more is ____. This means that we would expect a mean reading rate of 94.4

or higher from a population whose mean reading rate is 92 in ____ of every 100 random samples of size n=20 students. The new program is not abundantly more effective than the old program.

B. A mean reading rate of 94.4 wpm is unusual since the probability of obtaining a result of 94.4 wpm or more is ____. This means that we would expect a mean reading rate of 94.4

or higher from a population whose mean reading rate is 92 in ____ of every 100 random samples of size n=20 students. The new program is abundantly more effective than the old program.

Answer 1

µ =    92                                      
σ =    10                                      
n=   20                                      
X =   94.4
                                          
Z =   (X - µ )/(σ/√n) =(94.4-92)/(10/sqrt(20))=1.07

Using excel command =normsdist(1.07)

P(X ≥   94.4)= 0.1423

A. A mean reading rate of 94.4 wpm is not unusual since the probability of obtaining a result of 94.4 wpm or more is 0.1423. This means that we would expect a mean reading rate of 94.4

or higher from a population whose mean reading rate is 92 in 14 of every 100 random samples of size n=20 students. The new program is not abundantly more effective than the old program.