For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.79 million dollars. Assuming a population standard deviation gross earnings of 0.47 million dollars, obtain a 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions). Confidence interval: (_. b) Which of the following is the correct interpretation for your answer in part (a)? (A) We can be 99% confident that the mean gross earnings for this sample of 30 Rolling Stones concerts lies in the interval (B) We can be 99% confident that the mean gross earnings of all Rolling Stones concerts ies in the interval (C) There is a 99% chance that the mean gross earnings of all Rolling Stones concerts lies in the interval (D) None of the above

Respuesta :

Answer: a)  (2.57, 3.01)

b)  B) We can be 99% confident that the mean gross earnings of all Rolling Stones concerts ies in the interval

Step-by-step explanation:

a) Confidence interval formula :

[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]

, where n = sample size .

[tex]\sigma[/tex] = Population standard deviation.

[tex]\overline{x}[/tex] = Sample mean

[tex]z_{\alpha/2}[/tex] = Two-tailed z-value for significance level ([tex]\alpha[/tex]).

As per given :

n= 30

[tex]\overline{x}=2.79[/tex]

[tex]\sigma=0.47[/tex]

[tex]\alpha=1-0.99=0.01[/tex]

Using z-value table ,

Two-tailed z-value : [tex]z_{\alpha/2}=z_{0.005}=2.576[/tex]

Confidence interval :

[tex]2.79\pm (2.576)\dfrac{0.47}{\sqrt{30}}[/tex]

[tex]2.79\pm (2.576)(0.0858)[/tex]

[tex]\approx2.79\pm 0.22\\\\=(2.79-0.22,\ 2.79+0.22)=(2.57,\ 3.01)[/tex]

Thus , a 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions) : (2.57, 3.01)

Interpretation of a 99% confidence interval : We are 99% confident that the true population mean lies in it.

Thus , the correct interpretation of the given situation :

We can be 99% confident that the mean gross earnings of all Rolling Stones concerts lies in the interval.

Using the z-distribution, it is found that the 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions) is (2.57, 3.01).

The interpretation is:

(B) We can be 99% confident that the mean gross earnings of all Rolling Stones concerts lies in the interval.

We are given the standard deviation for the population, which is why the z-distribution is used to solve this question. We also have that:

  • The sample mean is of [tex]\overline{x} = 2.79[/tex].
  • The population standard deviation is of [tex]\sigma = 0.47[/tex].
  • The sample size is of [tex]n = 30[/tex].

The confidence interval is:

[tex]\overline{x} \pm z\frac{\sigma}{\sqrt{n}}[/tex]

The first step is finding the critical value, which is z with a p-value of [tex]\frac{1 + \alpha}{2}[/tex], in which [tex]\alpha[/tex] is the confidence level.

In this problem, [tex]\alpha = 0.99[/tex], thus, z with a p-value of [tex]\frac{1 + 0.95}{2} = 0.995[/tex], which means that it is z = 2.575.

Thus, the interval is given by:

[tex]\overline{x} - z\frac{\sigma}{\sqrt{n}} = 2.79 - 2.575\frac{0.47}{\sqrt{30}} = 2.57[/tex]

[tex]\overline{x} + z\frac{\sigma}{\sqrt{n}} = 2.79 + 2.575\frac{0.47}{\sqrt{30}} = 3.01[/tex]

The 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions) is (2.57, 3.01).

The interpretation is that we are 99% sure that the mean for all shows is in this interval, thus option B.

A similar problem is given at https://brainly.com/question/22596713