Question

A researcher compares the effectiveness of two different instructional methods for teaching physiology. A sample of 178 students using Method 1 produces a testing average of 54.4. A sample of 226 students using Method 2 produces a testing average of 74.5. Assume the standard deviation is known to be 18.58 for Method 1 and 9.52 for Method 2. Determine the 90% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.
Required:
a. Find the critical value that should be used in constructing the confidence interval.
b. Construct the 95% confidence interval. Round your answers to one decimal place.

Answer 1

Answer:

critical value = 1.645

The 90% confidence interval = ( -22.62, -17.58)

Step-by-step explanation:

Given that:

the sample  size $n_1$ = 178

the sample size $n_2$ = 226

the sample mean $\overline x_1$ = 54.4

the sample mean $\overline x_2$ = 74.5

population standard deviation $\sigma_1$ = 18.58

population standard deviation $\sigma_2$ = 9.52

level of significance ∝ = 1 - 0.90 = 0.10

The critical value for $Z_{\alpha/2} = Z _{0.10/2} = Z_{0.005}$ is 1.645

For the construction of our confidence interval, we use 90% since that is used to find the critical value.

∴

The margin of error = $Z \times\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}$

$1.645 \times\sqrt{\dfrac{18.58^2}{178} + \dfrac{9.52^2}{226}}$

$1.645 \times\sqrt{\dfrac{345.2164}{178} + \dfrac{90.6304}{226}}$

$1.645 \times\sqrt{2.34042}$

$\simeq$ 2.52

The lower limit = $( \overline x_1 - \overline x_2) - (M.O.E)$

= ( 54.4-74.5) - (2.52)

= -20.1 - 2.52

= -22.62

The upper limit = $( \overline x_1 - \overline x_2) + (M.O.E)$

= ( 54.4-74.5) + (2.52)

= -20.1 + 2.52

= -17.58

The 90% confidence interval = ( -22.62, -17.58)