For its study, 195 randomly selected Southerners were surveyed and found to watch a mean of 4.5 hours of sports per week. In the North, 148 randomly selected people were surveyed and found to watch a mean of 4.7 hours of sports per week. Find a 99% confidence interval for the true difference between the mean numbers of hours of sports watched per week for the two regions if the South has a population standard deviation of 1.5 hours per week and the North has a population standard deviation of 1.7 hours per week.

The 99% confidence interval for the true difference between the mean numbers of hours of sports watched per week for the two regions is = [ CI [-0.254, 0.654].

Step-by-step explanation:

The confidence interval for the true difference between the mean formula is given as:

μ1 - μ2 ± z × √(σ²1/n1 + σ²2/b2)

μ1 = Mean 1 = 4.5 hours

μ2 = Mean 2 = 4.7 hours

z = z score for 99% confidence interval = 2.576

σ1 = Standard deviation 1(South) = 1.5 hours

σ2 = Standard deviation 2(North) = 1.7 hours

n1 = number of samples 1 = 195

n2 = number of samples 2 = 148

Confidence Interval

= 4.5 - 4.7 ± 2.576 × √1.5²/195 + 1.7²/148

= 0.2 ± 2.576 × √0.0115384615 + 0.019527027

= 0.2 ± 2.576 × √0.0310954885

=0.2 ± 0.4542495969

Confidence Interval

= 0.2 - 0.4542495969

- 0.2542495969

= -0.254

Approximately = -0.654

= 0.2 + 0.4542495969

= 0.6542495969

Approximately = 0.654

Hence, The 99% confidence interval for the true difference between the mean numbers of hours of sports watched per week for the two regions is = CI [-0.254, 0.654].]

step-by-step explanation:

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)