Question

A camp director is interested in the mean number of letters each child sends during his or her camp session. The

population standard deviation is known to be $2.5$. A survey of $20$campers is taken. The mean from the sample is $7.9$with a

sample standard deviation of $2.8.$

a.

localid="1651507797100" $$\begin{gathered} i.x̄=\_\_\_\_\_\_\_\_ \\ ii.\sigma =\_\_\_\_\_\_\_\_ \\ iii.n=\_\_\_\_\_\_\_\_ \end{gathered}$$

b. Define the random variables $XandX̄$

in words.

c. Which distribution should you use for this problem? Explain your choice.

d. Construct a $90\%$confidence interval for the population mean number of letters campers send home.

i. State the confidence interval.

ii. Sketch the graph.

iii. Calculate the error bound.

e. What will happen to the error bound and confidence interval if $500$campers are surveyed? Why?

Short answer

(a) The result of part a is

$1.\overline{x}=7.9$

$2.\sigma =2.5$

$3.n=20$

(b) During the camp term, a single camper will send home localid="1651507776377" $X$letters, and the average number of letters sent home by $20$campers will be localid="1651507782181" $\overline{X}$.

(c) The problem is solved using the normal distribution. As we all know, the sample size for the population's standard deviation is greater than localid="1651507815140" $30,$and the distribution is localid="1651507820489" $N\left(7.9\begin{array}{c}2.9\\ \sqrt{20}\end{array}\right)$.

(d) the result of part d are

localid="1651507825330" $1.CI=(6.9805,8.8195)$

2. Through diagram

$6.9805$ $8.8195$

3. by calculating all

E B M=0.9195 .

(e)The confidence interval and error bound will decreases as we see all above .

Step-by-step solution

Explanation (a)

i. The average number of letters sent by $20$participants over the course of a camp session is $7.9$, $\overline{x}=7.9$

ii. The population standard deviation of the number of letters sent by campers throughout a camp session is, $\sigma =2.5$

iii. Size of the campers,

$n=20$

Explanation (b)

During the camp term, a single camper will send home $X$letters, and the average number of letters sent home by $20$campers will be $\overline{X}$.

Explanation (c)

The problem is solved using the normal distribution. As we all know, the sample size for the population's standard deviation is greater than $30$, and the distribution is $N\left(7.9\begin{array}{c}2.9\\ \sqrt{20}\end{array}\right)$.

Explanation (d) 

i. The confidence interval should be stated.

The confidence interval's output,

$n=20$

$CI=(6.9805,8.8195)$

ii. The graph is as follows:

$6.9805$ $8.8195$

iii. The formula is used to compute the error bound

$EBM=\frac{U_{\text{pper limit-lower limit}}}{2}$

$EBM=\frac{8.8195-6.9805}{2}$

$EBM=0.9195$.

Explanation (e)

The confidence interval and error bound will decreases as we see all above .