Question

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.

Answer 1

The confidence interval is
`-11.34 < \mu_1 -\mu_2 < -10.86`

Explanation:

From the question we are told that

The first sample size is
`n_1 = 146`

The second sample size is
`n_2 = 180`

The first sample mean is
`\= x_1 = 51.6`

The second sample mean is
`\= x_2 = 62.7`

The first standard deviation is
`\sigma _1 = 9.42`

The second standard deviation is
`\sigma _2 = 14.5`

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

`\alpha = (100 -98 )\%`

`\alpha = 2 \%`

`\alpha = 0.02`

Next we obtain the critical value of
`(\alpha )/(2)` from the z-table , the value is
`Z_{(\alpha )/(2) } = 2.33`

The reason we are obtaining critical value of

`(\alpha )/(2)`

instead of

`\alpha`

is because

`\alpha`

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

`1-\alpha`

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

`(\alpha )/(2)`

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

`E = Z_{(\alpha )/(2) } * \sqrt{ (\sigma_1^2)/(n_1^2) + (\sigma_2^2)/(n_2^2) }`

substituting values

`E = 2.33 * \sqrt{ (9.42^2)/(146^2) + (14.5^2)/(180^2) }`

substituting values

`E = 2.33 * \sqrt{ (9.42^2)/(146^2) + (14.5^2)/(180^2) }`

`E = 0.2405`

The 98% confidence interval is mathematically represented as

`(\= x _ 1 - \= x_2 ) - E < \mu_1 -\mu_2 < (\= x _ 1 - \= x_2 ) + E`

substituting values

`(51.6 - 62.7) - 0.2405 < \mu_1 -\mu_2 < (51.6 - 62.7) + 0.2405`

`-11.34 < \mu_1 -\mu_2 < -10.86`