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Can someone point me in the right direction with this Statistics problem? Any help at all is appreciated!? - Richarddddddd - 04-28-2014 04:00 AM
Research indicates that college students spend on average 4 hours on Facebook every
day. A random sample of 100 college students in Connecticut indicates that students
spend 3.5 hours on Facebook with a standard deviation of 2. Test the hypothesis that
college students in Connecticut spend less time on Facebook than the national average
(critical value is 1.645 when df ≥ 100).
- Charles D - 04-28-2014 04:01 AM
Ho: mu >=4
Ha: mu < 4 (claim)
x-bar = 3.5
s = 2
n = 100
test statisitc z = (x-mu)/ (s/sqrtn)) = ( 3.5 - 4 ) / (2 / sqrt(100) ) = -0.5 / 0.2 = - 2.5
The the null hypothesis has a <, the critical value is -1.645. The test statistic is well into the rejection region. So we reject Ho and there is sufficient evidence to accept the claim.
(See also "Khan Academy Hypothesis Testing")