Question

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.)

Answer 1

A) SE = 0.1118

B) 95% Confidence interval is;

-2.22 < μ1 - μ2 < -1.78

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