4.3 A college counselor is interested in estimating how many credits a student typically enrolls in each semester. The counselor decides to randomly sample 100 students by using the registrar’s database of students. The histogram below shows the distribution of the number of credits taken by these students. Sample statistics for this distribution are also provided.

Picture

Picture

a. What is the point estimate for the average number of credits taken per semester by students at this college? What about the median?

Mean: 13.65.

Median: 14.

b. What is the point estimate for the standard deviation of the number of credits taken per semester by students at this college? What about the IQR?

SD: 1.91.

IQR: 15 - 13 = 2.

c. Is a load of 16 credits unusually high for this college? What about 18 credits? Explain your reasoning. Hint: Observations farther than two standard deviations from the mean are usually considered to be unusual.

# 16 Credits

x <- 16
mu <- 13.65
SD <- 1.91
z <- (x - mu)/SD
z
## [1] 1.230366

Since z = 1.23 and |z| is less than 2, we conclude that 16 credits are not considered unusual.

#18 Credits 
x <- 18
mu <- 13.65
SD <- 1.91
z <- (x - mu)/SD
z
## [1] 2.277487

Since z = 2.28 and |z| is more than 2, we conclude that 18 credits are considered unusual.

d. The college counselor takes another random sample of 100 students and this time finds a sample mean of 14.02 units. Should she be surprised that this sample statistic is slightly different than the one from the original sample? Explain your reasoning.

Due to the Counselor is taking a sample which is an approximate represantation of the entire population (which will very from sample to sample),counselor should not be surpriesed.

e. The sample means given above are point estimates for the mean number of credits taken by all students at that college. What measures do we use to quantify the variability of this estimate(Hint: recall that \({ SD }_{ \overline { x } }=\frac { \sigma }{ \sqrt { n } }\) )? Compute this quantity using the data from the original sample.

We use standard error to quantify the variability.

\(SE=\frac { \sigma }{ \sqrt { n } } =\frac { 1.91 }{ 10 } =0.191\)