Question

The National Sleep Foundation surveyed representative samples of adults in six different countries to ask questions about sleeping habits.† Each person in a representative sample of 250 adults in each of these countries was asked how much sleep they get on a typical work night. For the United States, the sample mean was 391 minutes, and for Mexico the sample mean was 426 minutes. Suppose that the sample standard deviations were 26 minutes for the U.S. sample and 41 minutes for the Mexico sample. The report concludes that on average, adults in the United States get less sleep on work nights than adults in Mexico. Is this a reasonable conclusion? Support your answer with an appropriate hypothesis test. (Use α = 0.05. Use μ₁ for Mexico and μ₂ for the United States.)

Answer 1

Given that,
mean(x)=391
standard deviation , s.d1=26
number(n1)=250
y(mean)=426
standard deviation, s.d2 =41
number(n2)=250
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.651
since our test is left-tailed
reject Ho, if to < -1.651
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =391-426/sqrt((676/250)+(1681/250))
to =-11.3988
| to | =11.3988
critical value
the value of |t α| with min (n1-1, n2-1) i.e 249 d.f is 1.651
we got |to| = 11.39878 & | t α | = 1.651
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:left tail - Ha : ( p < -11.3988 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -11.3988
critical value: -1.651
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that on average, adults in the United States get less sleep on work nights than adults in Mexico.