Question

Twenty-five juniors from a University volunteer to allow their Math GRE test scores to be used in a study. These 25 juniors had a mean Math GRE score of 450. Suppose we know that the standard deviation of the population of Math GRE scores for juniors at the University is σ = 100. Assuming the population of Math GRE scores for juniors at the University is approximately Normally distributed, a 90% confidence interval for the mean Math GRE score μ for the population of juniors computed from these data is

Answer 1

The 90% confidence interval for the mean Math GRE score for the population of juniors at the university is between 417.1 and 482.9, calculated based on the provided mean, standard deviation, and sample size.

Step-by-step explanation:

To calculate the 90% confidence interval for the mean Math GRE score μ for the population of juniors, we can use the formula for a confidence interval when the population standard deviation σ is known:

Confidence interval = μ ± (Z * (σ / √N))

Where:

• μ = mean Math GRE score = 450
• σ = population standard deviation = 100
• N = sample size = 25
• Z = Z-value from the standard normal distribution for 90% confidence

First, we find the Z-value for a 90% confidence interval using a Z-table or calculator, usually approximately 1.645 for a 90% confidence level. Then, we substitute the known values into the formula:

Z * (σ / √N) = 1.645 * (100 / √25) = 1.645 * (100 / 5) = 1.645 * 20 = 32.9

Now we calculate the confidence interval:

Confidence interval = 450 ± 32.9

Which gives us:

Lower limit = 450 - 32.9 = 417.1

Upper limit = 450 + 32.9 = 482.9

Therefore, we estimate with 90 percent confidence that the true population mean Math GRE score for all juniors is between 417.1 and 482.9.

Answer 2

Let's start first by writing down the given:
σ = 100
sample mean = 450
sample size = 25

These information, plus the fact that we know that the population is approximately normally distributed, would tell us that we can use the normal distribution curve in analyzing the problem.

A confidence interval of the mean is just a range statistically estimated to contain the population mean. For a 90% confidence interval, we would look at the Z-table and see where 90% of the data falls. We'll notice that it will fall within 1.645 standard deviations of the mean.

Next, we look for the standard error of the mean. This will have a formula

`error= (standard deviation of population)/( √(N) )`

The standard error would just therefore be equal to

`(100)/( √(25) )= (100)/(5) =20`

Lastly, we just get the product of the standard error and 1.645 and add it to 450 for the maximum value and subtract it to 450 for the minimum value.

`450-20(1.645)=417.1`

`450+20(1.645)=482.9`

ANSWER: 417.1<μ<482.9