11. In a study conducted to determine the role that sleep disorders play in acad

Question

11. In a study conducted to determine the role that sleep disorders play in academic performance, researcher Jane Gaultney conducted a survey of 1845 college students. The students completed a survey to determine if they had a sleep disorder (such as narcolepsy, insomnia, or restless leg syndrome). Of the 503 students with a sleep disorder, the mean grade point average was 2.65 with a standard deviation of 0.87. Of the 1342 students without a sleep disorder, the mean grade point average was 2.82 with a standard deviation of 0.83. Source: SLEEP 2010: Associated Professional Sleep Societies 24th Annual Meeting

Construct a 95% confidence interval for the difference between the students GPA without a sleep disorder and the students GPA with a sleep disorder. (Hint: Use 2-SampTInt) Round 2 decimal places.

Interpret the interval

Is there evidence to suggest sleep disorders adversely affect one's GPA at the = 0.05 level of significance?

What is the null and alternative hypothesis?

Calculate the P-value (Use 2-SampTTest on the TI Calculator and remember the P-value is p)

Round 3 decimals

Compare the P-value to =0.05 level of significance. Is the p-value less than, greater than, or equal to the level of significance = 0.05?

What is the result of the test? Reject H0 or Do Not Reject H0?

State the conclusion.

Explanation / Answer

Q1.
given that,
mean(x)=2.65
standard deviation , s.d1=0.87
number(n1)=503
y(mean)=2.82
standard deviation, s.d2 =0.83
number(n2)=1342
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((0.757/503)+(0.689/1342))
= 0.045
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 502 d.f is 1.965
margin of error = 1.965 * 0.045
= 0.088
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (2.65-2.82) ± 0.088 ]
= [-0.258 , -0.082]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=2.65
standard deviation , s.d1=0.87
sample size, n1=503
y(mean)=2.82
standard deviation, s.d2 =0.83
sample size,n2 =1342
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 2.65-2.82) ± t a/2 * sqrt((0.757/503)+(0.689/1342)]
= [ (-0.17) ± t a/2 * 0.045]
= [-0.258 , -0.082]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-0.258 , -0.082] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

Q2.

Given that,
mean(x)=2.65
standard deviation , s.d1=0.87
number(n1)=503
y(mean)=2.82
standard deviation, s.d2 =0.83
number(n2)=1342
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =1.965
since our test is two-tailed
reject Ho, if to < -1.965 OR if to > 1.965
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =2.65-2.82/sqrt((0.7569/503)+(0.6889/1342))
to =-3.7842
| to | =3.7842
critical value
the value of |t | with min (n1-1, n2-1) i.e 502 d.f is 1.965
we got |to| = 3.78422 & | t | = 1.965
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -3.7842 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -3.7842
critical value: -1.965 , 1.965
decision: reject Ho
p-value: 0

we have evidence to suggest sleep disorders adversely affect one's GPA

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