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

Answer:

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)

Step-by-step 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 $\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$, i.e., $\sqrt{\frac{(14.26)^{2}}{181}+\frac{(7.48)^{2}}{227}}$ = 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

Answer 2

Answer:

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)