Two shooters, Rodney and Philip, practice at a shooting range. They fire 100 rounds each at separate targets. The targets are marked with circles and each bullet hitting a particular circle gets them a particular number of points. 60 rounds are selected at random. The sample mean scores of Rodney and Philip are 8 and 10, respectively. And, their variances are 0.25 and 0.50, respectively The standard error of the difference between their mean scores is (Round your answer to two decimal places.) The 95% confidence interval for the difference between the mean scores Rodney and Philip is : (Enter a minus sign if your answer is negative. Round your answer to two decimal places.)

A) SE = 0.1118

B) 95% Confidence interval is;

-2.22 < μ1 - μ2 < -1.78

Step-by-step explanation:

We are given;

Sample size; n = 60

Sample mean 1; x1 = 8

Sample mean 2; x2 = 10

Variance 1; v1 = 0.25

Variance 2; V2 = 0.5

Now,

Standard deviation is given by the formula ;

σ = √variance

Thus;

σ1 = √0.25

σ1 = 0.5

σ2 = √0.5

σ2 = 0.7071

Formula for standard error of mean is;

SE = σ/√n

Thus;

SE1 = 0.5/√60

SE1 = 0.06455

SE2 = 0.7071/√60

SE2 = 0.0913

The standard error of the difference between their mean scores would be;

SE = √[(SE1)² + (SE2)²]

SE = √(0.06455² + 0.0913²)

SE = 0.1118

Formula for 95% confidence interval is;

(x1 - x2) - ((Z_α/2) × SE) < μ1 - μ2 < (x1 - x2) + ((Z_α/2) × SE)

Where Z_α/2 is the critical value. At confidence interval of 95%, we have a critical value of 1.96 from tables.

Thus, plugging in the relevant values, we have;

(8 - 10) - (1.96 × 0.1118) < μ1 - μ2 < (8 - 10) + (1.96 × 0.1118)

-2.22 < μ1 - μ2 < -1.78

the answer is d

the answer is in the first quadrant, which make the y-int positive.

since the dots are going down, therefore the correlation is negative

i hope it is right.

c: lkllmn.

hope it's you