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.

i. $\overline{x}$=________

ii. σ =________

iii. n =________

b. Define the random variables X and $\overline{x}$in words.

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

d. Construct a $90\%$confidence interval for the population mean the 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) i. $$\begin{gathered} \overline{x}=7.9 \\ \sigma =2.5 \\ n=20 \end{gathered}$$

(b) The average number of letters sent home by $20$campers during the camp session is $\overline{X}$, while the number of letters sent home by a single camper is $\overline{X}$.

(c) The distribution of the problem is $N(7.9,\frac{2.5}{\sqrt{20}})$

(d) i) $CI=(6.9805,8.8195)$

ii)

iii) $EBM=0.9195$

(e) Decrease.

Step-by-step solution

Part (a) Step 1: Concept introduction

The standard deviation is a way to gauge what further fluctuation or divergence overall is in a sequence of data measurements.

The mean of a time series seems to be the typical (centre), and the beta value is the scatter (differential) of results all around the norm.

Part (a) Step 2: Explanation

i. The sample mean number of letters sent by $20$campers throughout the camp session is $7.9$,

$\overline{x}=7.9$

ii. The population standard deviation of the number of letters sent by campers during the camp session is,

$\sigma =2.5$

iii. The size of the campers is,

Part (b) Step 1: Concept introduction

The standard deviation is a way to gauge what further fluctuation or divergence overall is in a sequence of data measurements.

The mean of a time series seems to be the typical (centre), and the beta value is the scatter (differential) of results all around the norm.

Part (b) Step 2: Explanation

The average number of letters sent home by $20$campers during the camp session is $\overline{X}$, while the number of letters sent home by a single camper is$\overline{X}$.

Part (c) Step 1: Concept introduction

The standard deviation is a way to gauge what further fluctuation or divergence overall is in a sequence of data measurements.

The mean of a time series seems to be the typical (centre), and the beta value is the scatter (differential) of results all around the norm.

Part (c) Step 2: Explanation

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$N(7.9,\frac{2.5}{\sqrt{20}})$.

Part (d) Step 1: Concept introduction

The standard deviation is a way to gauge what further fluctuation or divergence overall is in a sequence of data measurements.

The mean of a time series seems to be the typical (centre), and the beta value is the scatter (differential) of results all around the norm.

Part (d) Step 2: Explanation

i) The confidence interval is

CI $=6.9805,8.8195$

ii) The graph is

iii) $EBM=\frac{\text{Upper limit-lower limit}}{2}$

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

$=0.9195$

Part (e) Step 1: Concept introduction

The standard deviation is a way to gauge what further fluctuation or divergence overall is in a sequence of data measurements.

The mean of a time series seems to be the typical (centre), and the beta value is the scatter (differential) of results all around the norm.

Part (e) Step 2: Explanation

The confidence interval and error bound will shrink.