Question

A researcher 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.

Answer 1

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

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)

Answer 2

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-10.2227, -4.7773)

Explanation:

Let
`\mu_(1)-\mu_(2)` be the true difference between testing averages for students using Method 1 and students using Method 2. We have the large sample sizes
`n_(1) = 181` students using Method 1 and
`n_(2) = 227` students using Method 2, the unbiased point estimate for
`\mu_(1)-\mu_(2)` is
`\bar{x}_(1) - \bar{x}_(2)`, i.e., 57.5 - 65 = -7.5

The standard error is given by
`\sqrt{(\sigma^(2)_(1))/(n_(1))+(\sigma^(2)_(2))/(n_(2))}`, i.e.,
`\sqrt{((14.26)^(2))/(181)+((7.48)^(2))/(227)}` = 1.1704.

Then, the endpoints for a 98% confidence interval for
`\mu_(1)-\mu_(2)` is given by

-7.5-
`(z_(0.02/2))`1.1704 and -7.5+
`(z_(0.02/2))`1.1704, i.e.,

-7.5-
`(z_(0.01))`1.1704 and -7.5+
`(z_(0.01))`1.1704 where
`z_(0.01)` is the 1st quantile of the standard normal distribution, i.e., -2.3263, so, we have

-7.5-(2.3263)(1.1704) and -7.5+(2.3263)(1.1704), i.e.,

-10.2227 and -4.7773