A researcher compares the effectiveness of two different instructional methods for teaching pharmacology. A sample of 63 students using Method 1 produces a testing average of 52.2. A sample of 93 students using Method 2 produces a testing average of 72.3. Assume that the population standard deviation for Method 1 is 15.92, while the population standard deviation for Method 2 is 17.96. Determine the 95% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.

The 95% confidence interval for the difference between means is (-25.5, -14.7).

Step-by-step explanation:

We have to calculate a 95% confidence interval for the difference between means.

The sample 1 (Method 1), of size n1=63 has a mean of 52.2 and a standard deviation of 15.92.

The sample 2 (Method 2), of size n2=93 has a mean of 72.3 and a standard deviation of 17.96.

The difference between sample means is Md=-20.1.

The estimated standard error of the difference between means is computed using the formula:

The critical t-value for a 95% confidence interval is t=1.975.

The margin of error (MOE) can be calculated as:

Then, the lower and upper bounds of the confidence interval are:

The 95% confidence interval for the difference between means is (-25.5, -14.7).

mr. flanders’ class sold 47 doughnuts and   mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations:

more detail

step-by-step explanation: