Question

A random sample of 70 observations from a normally distributed population possesses a sample mean equal to 26.2 and a sample standard deviation equal to 4.1.

a. Find an approximate 95% confidence interval for

b. What do you mean when you say that a confidence coefficient is .95?

c. Find an approximate 99% confidence interval for

d. What happens to the width of a confidence interval as the value of the confidence coefficient is increased while the sample size is held fixed?

e. Would your confidence intervals of parts a and c be valid if the distribution of the original population was not normal? Explain

Short answer

A confidence interval is described as the set of numbers seen in our sample for which we anticipate discovering the number that best represents the overall population.

Step-by-step solution

Step-by-Step Solution Step 1: (a) The data is given below

$\text{Samplemean}\stackrel{-}{\chi }\text{=26.2}$

The sample size is high; the sample standard deviation can be used to approximate the overall standard deviations.

$\begin{array}{l}\sigma =4.1\\ \text{n}=70\end{array}$

The calculation is given below:

$$\begin{gathered} \text{Levelofsignificance=}\alpha =1-\text{Confidence} \\ =1-0.95 \\ =0.05 \end{gathered}$$

$$\begin{gathered} \text{Criticalvalue=z}\alpha /2 \\ ={\text{z}}_{0.025}=1.96 \end{gathered}$$

$$\begin{gathered} \text{Criticalvalue,}\pm \text{z}\alpha /2=\pm 1.96 \\ \text{Errorofmargin}\left(\text{E}\right)\text{=z}\alpha /2\times \frac{\sigma }{\sqrt{\text{n}}} \\ \text{E=1.962×}\frac{\text{4.1}}{\sqrt{\text{70}}} \\ \text{E=0.960486} \end{gathered}$$

95% confidence interval limits are shown by:

$$\begin{gathered} \text{Lowerlimit=}\stackrel{-}{\chi }\text{-E} \\ =26.2-0.960486 \\ =25.2395 \end{gathered}$$

$$\begin{gathered} \text{Upperlimit=}\stackrel{-}{\chi }+\text{E} \\ =26.2-0.960486 \\ =27.1605 \end{gathered}$$

95% confidence interval is:

$=\stackrel{-}{\chi }+\text{E}$

localid="1651472418180" $$\begin{gathered} \text{=26.2}\pm \text{0.960486} \\ =(25.239514,27.160486) \end{gathered}$$

(b) Confidence coefficient

Since the approach we used performs 95% of the period, we have a confidence coefficient of 0.95that the calculated confidence interval will include the population's mean.

(c) The data is given below

$\text{Samplemean}\stackrel{-}{\chi }\text{=26.2}$

$\begin{array}{l}\sigma =4.1\\ \text{n}=70\end{array}$

The calculation is given below:

$$\begin{gathered} \text{Levelofsignificance=}\alpha =1-\text{Confidence} \\ =1-0.99 \\ =0.01 \end{gathered}$$

$$\begin{gathered} \text{Criticalvalue=z}\alpha /2 \\ ={\text{z}}_{0.005}=2.58 \end{gathered}$$

The closest value from the z table:

$$\begin{gathered} \text{Criticalvalue,}\pm \text{z}\alpha /2=\pm 2.58 \\ \text{Errorofmargin}\left(\text{E}\right)\text{=z}\alpha /2\times \frac{\sigma }{\sqrt{\text{n}}} \\ \text{E=2.58×}\frac{\text{4.1}}{\sqrt{\text{70}}} \\ \text{E=1.264314} \end{gathered}$$

95% confidence interval limits are shown by:

$$\begin{gathered} \text{Lowerlimit=}\stackrel{-}{\chi }\text{-E} \\ =26.2-1.264314 \\ =24.9357 \end{gathered}$$

$$\begin{gathered} \text{Upperlimit=}\stackrel{-}{\chi }+\text{E} \\ =26.2-1.264314 \\ =27.4643 \end{gathered}$$

95% confidence interval is:

localid="1651473127163" $$\begin{gathered} \stackrel{-}{\chi }+\text{E} \\ \text{=26.2}\pm 1.2\text{64314} \\ =(24.935686,27.464314) \end{gathered}$$

(d) Confidence interval width

If the confidence coefficient is raised, the width increases.

(e) If the initial population's distribution was not normal, confidence intervals for parts an as well as c would be valid

Yes, it will be validbecause of the central limit theorem, sample means are about distributed normally when the sample size is larger than 30.