A researcher compares the effectiveness of two different instructional methods for teaching anatomy. A sample of 146 students using Method 1 produces a testing average of 51.6. A sample of 180 students using Method 2 produces a testing average of 62.7. Assume the standard deviation is known to be 9.42 for Method 1 and 14.5 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. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval.

The  confidence interval is  

Step-by-step explanation:

From the question we are told that

    The first sample size is  

    The second sample size is

    The first sample mean is

    The second  sample  mean is  

     The first standard deviation is  

     The second standard deviation is  

Given that the confidence level is  98% then the significance level is mathematically evaluated as

Next we obtain the critical value of   from the z-table , the value is  

The reason we are obtaining critical value of  

instead of  

is because  

represents the area under the normal curve where the confidence level interval (

) did not cover which include both the left and right tail while  

is just the area of one tail which what we required to calculate the margin of error

NOTE: We can also obtain the value using critical value calculator (math dot armstrong dot edu)

Generally the margin of error is mathematically represented as

substituting values

substituting values

The 98% confidence interval is mathematically represented as

substituting values

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116