Aresearcher compares the effectiveness of two different instructional methods for teaching physiology. a sample of 181 students using method 1 produces a testing average of 57.5. a sample of 227 students using method 2 produces a testing average of 65. assume the standard deviation is known to be 14.26 for method 1 and 7.48 for method 2. determine the 98% confidence interval for the true difference between testing averages for students using method 1 and students using method 2.

98% confidence interval using Method 1 is (55.01, 59.99) and Method 2 is (63.84, 66.16)

Step-by-step explanation:

Confidence Interval (CI) = mean + or - (t×sd)/√n

Method 1

Mean = 57.5, sd = 14.26, n = 181, degree of freedom = n - 1 = 181 - 1 = 180

t-value corresponding to 180 degrees of freedom and 98% confidence level is 2.3474

Lower bound = 57.5 - (2.3474 × 14.26)/√181 = 57.5 - 2.49 = 55.01

Upper bound = 57.5 + (2.3474 × 14.26)/√181 = 57.5 + 2.49 = 59.99

98% confidence interval using Method 1 is (55.01, 59.99)

Method 2

Mean = 65, sd = 7.48, n = 227, degree of freedom = n - 1 = 227 - 1 = 226

t-value corresponding to 226 degrees of freedom and 98% confidence level is 2.3434

Lower bound = 65 - (2.3434 × 7.48)/√227 = 65 - 1.16 = 63.84

Upper bound = 65 - (2.3434 × 7.48)/√227 = 65 - 1.16 = 6384

98% confidence interval using Method 2 is (63.84, 66.16)

what is the answer?

answer: 10 turtles

step-by-step explanation: as turtles have four legs so 4.10=40 therefore birds are 4 which sue counted