Question

Some researchers are interested in the GPA difference between freshmen and sophomores. The distributionsof GPAs in both groups are approximately normal. From previous school records, we know that themean and standard deviation for freshmen are 3.5 and 0.5, respectively. Similarly, the mean and standarddeviation for sophomores are 3.2 and 0.8, respectively. Independent random samples of 40 students areto be selected from both years (for a total of 80 students).
a. If the sample mean GPA is to be calculated for both groups and we calculate the difference between freshmen and sophomores, what is the expected value for the difference in sample means?
b. If the sample mean GPA is to be calculated for both groups and we calculate the difference between freshmen and sophomores, what is the standard deviation of the sampling distribution of the difference in sample means?
c. What is the probability that the average GPA in freshmen is higher than the average GPA in the sample of sophomores?

Answer 1

Answer:

a) $E(X-Y) = E(X)-E(Y) = \mu_x -\mu_y = 3.5-3.2=0.3$

b) $Var(\bar X -\bar Y) = \frac{\sigma^2_x}{n_x} +\frac{\sigma^2_y}{n_y}$

And replacing we got:

$Var(\bar X -\bar Y) =\frac{0.5^2}{40} + \frac{0.8^2}{40} = 0.02225$

And the deviation would be:

$sd(\bar X -\bar Y) = \sqrt{0.02225}= 0.149$

c) $P(\bar X >\bar Y)$

And we can use the z score formula given by:

$z = \frac{D-\mu_d}{\sigma_d}$

$d = X-Y \sim N (\mu_d =0.3, sigma_d = 0.149)$

And replacing we got:

$P(d >0) = P(z> \frac{\bar X- \bar Y -0}{\sigma_d}) =P(z> \frac{3.5-3.2}{0.149}) =P(z>2.103)$

And using the complement rule we got:

$P(z>2.103) = 1-P(z$

Step-by-step explanation:

For this case we have the following info:

Freshmem

$X \sim N(\mu= 3.5 , \sigma=0.5)$

Sophomores

$Y \sim N(\mu= 3.2 , \sigma=0.8)$

We select a sample of 40 for both groups $n_x =n_y =40$

Part a

The expected value for the difference is given by:

$E(X-Y) = E(X)-E(Y) = \mu_x -\mu_y = 3.5-3.2=0.3$

Part b

We know that the sample mean follows this distribution:

$\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})$

And we want the distribution for $\bar X-\bar Y$

And we need to find the variance with the following formula:

$Var(\bar X -\bar Y) = \frac{\sigma^2_x}{n_x} +\frac{\sigma^2_y}{n_y}$

And replacing we got:

$Var(\bar X -\bar Y) =\frac{0.5^2}{40} + \frac{0.8^2}{40} = 0.02225$

And the deviation would be:

$sd(\bar X -\bar Y) = \sqrt{0.02225}= 0.149$

Part c

For this case we want this probability:

$P(\bar X >\bar Y)$

And we can use the z score formula given by:

$z = \frac{D-\mu_d}{\sigma_d}$

$d = X-Y \sim N (\mu_d =0.3, sigma_d = 0.149)$

And replacing we got:

$P(d >0) = P(z> \frac{\bar X- \bar Y -0}{\sigma_d}) =P(z> \frac{3.5-3.2}{0.149}) =P(z>2.103)$

And using the complement rule we got:

$P(z>2.103) = 1-P(z$

Answer 2

Answer and Step-by-step explanation:

a)

expected value for the difference in sample means=3.5-3.2=0.3

b)

standard deviation of the sampling distribution of the difference in sample means

=sqrt((0.5^2/40)+(0.8^2/40))

=0.1492

c)

z=(0.3-0)/0.1492

z=2.01

P(z>2.01)=0.0222