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.

critical value = 1.645

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

Step-by-step explanation:

Given that:

the sample  size = 178

the sample size = 226

the sample mean = 54.4

the sample mean = 74.5

population standard deviation = 18.58

population standard deviation = 9.52

level of significance ∝ = 1 - 0.90 = 0.10

The critical value for 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 =

2.52

The lower limit =

= ( 54.4-74.5) - (2.52)

= -20.1 - 2.52

= -22.62

The upper limit =

= ( 54.4-74.5) + (2.52)

= -20.1 + 2.52

= -17.58

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

2,600 divided by 80 = 32.5

step-by-step explanation:

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8