Can someone point me in the right direction with this Statistics problem? Any help at all is appreciated!?
|
04-28-2014, 04:00 AM
Post: #1
|
|||
|
|||
Can someone point me in the right direction with this Statistics problem? Any help at all is appreciated!?
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). Ads |
|||
04-28-2014, 04:01 AM
Post: #2
|
|||
|
|||
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") Ads |
|||
« Next Oldest | Next Newest »
|
User(s) browsing this thread: 1 Guest(s)