The amount of precipitation (in inches) in June of a recent year was measured in some randomly selected Michigan and Ohio cities (see below). Assume that the mean amount of June precipitation in Michigan and Ohio cities are both approximately normally distributed.
Construct a 98% confidence interval for the difference of the mean amount of June precipitation in Michigan cities minus mean amount of June precipitation in Ohio cities.
Michigan Ohio
Lansing :3.46 Akron:3.15
BigRapids :3.27 Dayton:4.17
Monroe:3.62 Fremont:4.06
Marquette:2.68 Toledo:3.86
Alpena:2.68 Cincinnate:4.17

The 98% confidence interval for the difference of the mean amount of June precipitation in Michigan cities and Ohio cities is (-1.77, 0.29).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between two population mean is:

Compute the value of sample means and sample standard deviations from the provided data.

The critical value of t for 98% confidence level and (n - 1) = 4 degrees of freedom is:

Compute the 98% confidence interval for the difference of the mean amount of June precipitation in Michigan cities and Ohio cities as follows:

Thus, the 98% confidence interval for the difference of the mean amount of June precipitation in Michigan cities and Ohio cities is (-1.77, 0.29).

i would pic false cause it goes down towards negative

step-by-step explanation:

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm