Answer to Question #338202 in Statistics and Probability for Carrot

Question

A. Find the length of the confidence interval (s = standard deviation)

1. s = 3

n = 250

Confidence level = 95%

2. s = 6

n = 400

Confidence level = 99%

B. Determine the sample size, given the following data.

1. s = 5

E = 2.42

Confidence level = 95%

2. You want to estimate the mean gasoline price within your town to the margin of error of 6 centavos. Local newspaper reports the standard deviation for gas price in the area is 30 centavos. What sample size is needed to estimate the mean gas prices at 99% confidence level?

3. Carlos wants to replicate a study where the highest observed value is 14.8 while the lowest is 14.2. He wants to estimate the population mean µ to the margin of error of 0.025 of its true value. Using 95% confidence level, find the sample size n that he need.

Answer 1

A.

1. The critical value for $\alpha = 0.05$ and $df=n-1=249$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 1.969537.$

$$Length=2t_c\times\dfrac{s}{\sqrt{n}}=2(1.969537)\times\dfrac{3}{\sqrt{250}}$$

$$=0.7474$$

2. The critical value for $\alpha = 0.01$ and $df=n-1=399$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.588207.$

$$Length=2t_c\times\dfrac{s}{\sqrt{n}}=2(2.588207)\times\dfrac{6}{\sqrt{400}}$$

$$=0.7765$$

B.

1.

$$E=t_c\times\dfrac{s}{\sqrt{n}}$$

The critical value for $\alpha = 0.05$ and $df=n-1=18$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.100922.$

$$E=2.100922\times\dfrac{5}{\sqrt{19}}=2.41$$

The critical value for $\alpha = 0.05$ and $df=n-1=17$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.1098162.$

$$E= 2.109816\times\dfrac{5}{\sqrt{18}}=2.49$$

$n=19$

2. The critical value for $\alpha = 0.01$ is $z_c = z_{1-\alpha/2} = 2.5758.$

$$E=z_c\times\dfrac{\sigma}{\sqrt{n}}$$

$$n\ge(\dfrac{z_c\times\sigma}{E})^2$$

$$n\ge(\dfrac{2.5758\times30}{6})^2$$

$$n=166$$

3.

$$\bar{x}=\dfrac{14.2+14.8}{2}=14.5$$

$$E=0.025(14.5)=0.3625$$

The critical value for $\alpha = 0.05$ is $z_c = z_{1-\alpha/2} = 1.96.$

$$E=z_c\times\dfrac{\sigma}{\sqrt{n}}$$

$$n\ge(\dfrac{z_c\times\sigma}{E})^2$$

$$n\ge(\dfrac{1.96\times\sigma}{0.3625})^2$$

Let $\sigma=1$

$$n=30$$