Answer to Question #175352 in Statistics and Probability for Denisse Bisuña

Question

1.Given the following: the sampled population is normally distributed , n = 58, X = 75, and σ = 10.

a. What is the 99% confidence interval for μ?

b. Are the assumptions met? Explain your answer.

2.The ages of random sample of 60 Grade 9 students were obtained to estimate the mean age of all Grade 9 students. X = 15.3 years and the population variance is 16 years.

a. What is the point estimate for μ?

b. What is the 95% confidence interval for μ?

c. What is the 99% confidence interval for μ?

d. What conclusions can you make based on each estimate?

Answer 1

1.

a. The critical value for $\alpha=0.01$ is $z_c=z_{1-\alpha/2}=2.576.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{x}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$

$$=(75-2.576\times\dfrac{10}{\sqrt{58}}, 75+2.576\times\dfrac{10}{\sqrt{58}})$$

$$=(71.618, 78.382)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $71.618<\mu< 78.382,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(71.618, 78.382).$ 

b. There are three important conditions that we should check: Random, Normal, and Independent.

The confidence interval of the mean of a measurement variable is commonly estimated on the assumption that the statistic follows a normal distribution.

As a general rule, sample sizes equal to or greater than 30 are deemed sufficient for the CLT to hold, meaning that the distribution of the sample means is fairly normally distributed.

$n=58>30$

The assumptions are met.

2.

a. The point estimate for $\mu$ is  the sample mean $\bar{x}=15.3$ years.

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

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{x}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$

$$=(15.3-1.96\times\dfrac{16}{\sqrt{60}}, 15.3+1.96\times\dfrac{16}{\sqrt{60}})$$

$$=(11.251, 19.349)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $11.251<\mu< 19.349,$ which indicates that we are 95% confident that the true population mean $\mu$  is contained by the interval $(11.251, 19.349).$  

c. The critical value for $\alpha=0.01$ is $z_c=z_{1-\alpha/2}=2.576.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{x}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$

$$=(15.3-2.576\times\dfrac{16}{\sqrt{60}}, 15.3+2.576\times\dfrac{16}{\sqrt{60}})$$

$$=(9.979, 20.621)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $9.979<\mu< 20.621,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(9.979, 20.621).$ 

d. A 99% confidence interval would be wider than a 95 % confidence interval.

The width of the confidence interval will be larger when the confidence level is higher (because you can have greater confidence when you are less precise).

The width of a confidence interval will be smaller when you have a larger sample size (because larger samples make sample statistics more reliable).

The width of the confidence interval will be larger when the underlying population has a larger standard deviation (because more variability makes sample statistics less reliable).

The width of the confidence interval will be larger when the confidence level is higher (because you can have greater confidence when you are less precise).