Question

When people take sleeping medicine they would like to go to sleep as quickly as possible. For a new sleeping drug a study was done comparing time to go to sleep with the new pill to known time to go to sleep without any pill. Previous research shows that on average without a pill people fall asleep in mean time of μ{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>μ</mi></math>"} = 15 minutes with a standard deviation of σ = 10 minutes. A study was done on 40 individuals where the new sleeping pill was given to them and their time to fall asleep was recorded. The sample average

X¯{"version":"1.1","math":"\bar X"}

for this sample of 40 test subjects was recorded.

If the sleeping pill was having NO effect then we would expect the sample average to be close to 15 minutes. Now suppose the sample average for the subjects was 1 minute. That is, with the sleeping pill the average time to fall to sleep is 1 minute. Obviously we need no probability calculations to know that it would be very unlikely to get such a short time compared to the no pill mean of 15 minutes if the sleeping pill was not working. So without statistical calculations we would assume that the pill was having an effect. But what about if the sample mean of the 40 individuals was m= 10 minutes ? Now we need a probability calculation to judge if this could happen if the pill had no effect.

To judge the likelihood of such results the researchers calculate the probability that sample mean could be less than 14, 13, 12, 11, 10, and so on. If the probability of getting a sample mean of 10 or less is very low if the pill is not working then most would conclude that if 10 did occur then the pill is working.

Your problem is to calculate

P(X¯≤m){"version":"1.1","math":"P(\bar X \leq 10 )"}

where m = 10 . That is, find the probability P(sample mean is ≤  10 assuming that the pill has no effect) . Give your answer to 4 decimal places, do not round up or down.

Note you may assume the distribution of the variable x = lenght of time to go to sleep is normal , or you may assume that distribution of x is unknown but since sample size is large the Central Limit Theorem applies and again we may use the normal for our calculations.

Answer 1

We need to assume the distribution of null hypothesis to calculate the required probability, which is the Normal with,

We need to find, P(10) and n=40.

P(10)

=

=P(Z-3.1623)

=0.000783 {from z-table}

0.0007 {Since, it is written in the question: do not round up or down.}