According to the U. S. Energy Information Administration the average number of televisions per household in the United States was 2.3. A college student claims the average number of TV’s per household in the United States is different. He obtains a random sample of 73 households and finds the mean number of TV’s to be 2.1 with a standard deviation of 0.84. Test the student’s claim at the 0.01 significance level.

Let be the average number of televisions per household in the United States .

As per given ,

Since is two-tailed and population standard deviation is unknown, so the test is two-tailed t-test.

For sample : Sample size : n= 73, sample mean: = 2.1, sample standard deviation : s= 0.84.

T-critical value for degree of freedom n-1 = 73-1=72 and 0.01 significance level is 2.646 . [By students' t-distribution table]

Since, i.e.

This means we cannot reject null hypothesis.

We conclude that the average number of televisions per household in the United States is 2.3 at the 0.01 significance level.

do your own work and don"t cheat.

step-by-step explanation:

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)