In a study to investigate whether lack of sleep hurts short term memory, 100 adu
ID: 3320737 • Letter: I
Question
In a study to investigate whether lack of sleep hurts short term memory, 100 adults were each tested twice. Both times they studied a list of 30 words for two minutes, and they were then asked to recall as many as possible, with a different but comparable list used for each time. For one of the two test they had been without sleep for 24 hours, while for the other they have a normal amount of sleep. The order was randomly assigned. The difference in number of words recalled (with sleep vs without sleep) was recorded for each person. The average difference was 1.5 words, with a standard deviation of 6.1 words.
Carry out the five steps of the hypothesis test for this situation, filling in the details in the space provided. Use = 0.05. Step
1: Determine the null and alternative hypothesis (Be sure you define the parameter you use).
Step 2: Verify necessary data conditions, and if met, summarize the data into an appropriate test statistic.
Step 3: Assuming the null hypothesis is true, find the p-value.
Step 4: Decide whether the result is statistically significant.
Step 5: Report the conclusion in the context of the situation.
Explanation / Answer
Given that,
population mean(u)=30
standard deviation, =6.1
sample mean, x =28.5
number (n)=100
null, Ho: =30
alternate, H1: !=30
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 28.5-30/(6.1/sqrt(100)
zo = -2.459
| zo | = 2.459
critical value
the value of |z | at los 5% is 1.96
we got |zo| =2.459 & | z | = 1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -2.459 ) = 0.014
hence value of p0.05 > 0.014, here we reject Ho
ANSWERS
---------------
1.
null, Ho: =30
alternate, H1: !=30
2.
test statistic: -2.459
3.
critical value: -1.96 , 1.96
4.
p-value: 0.014
5.
decision: reject Ho
we have enough evidence to support the claim
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