Question

A schoolteacher is concerned that her students watch more TV than the average American child. She...

A schoolteacher is concerned that her students watch more TV than the average American child. She reads that according to the American Academy of Pediatrics (AAP), the average American child watches 4 hours of TV per day (μ = 4.0 hours). She records the number of hours of TV each of her six students watch per day. The times (in hours) are 2.6, 4.1, 5.4, 4.3, 2.8, and 4.8.

cohen's d d=

.05 level of significance and a one independent t test state the value of the test statistic

t=

H0:mu= 4 i.e.( Average American child watches 4 hr tv per day)

Vs

H1:mu >4 i.e.( claim of school teacher)

Under , the null hypothesis

Test statistic, t= x bar- mu/(s/√n) follows t distribution with n-1 degree of freedom.

X bar= 4.0 ( average of the given data)

std. Deviation= s= 1.1045

So, t= 4-4/(1.1045/√6)= 0

Critical t - value for alpha= 0.05 and for (6-1) df = 2.015

Since, critical t value is greater than calculated test statistic value so, we don't reject the null hypothesis and concluded that Average American watches 4 hr tv per day.

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