Answer to Question #350342 in Statistics and Probability for Jcanne

Question

1.The population of SULU Horn bill (one of the endangered bird species in the Philippines) has a standard deviation of 40. An environment researcher wants to construct a 90% confidence interval if the sample size is 150 and the sample mean is 65.

a. What is the margin of error

b. Construct the confidence interval

c. Find the leng of the confidence interval

2. The average price of 350 cellphones is Php 13,500 with a sample standard deviation of Php 750. A market researcher desires a 99% level of confidence in the true average price of cellphones

a. What is the margin error

b. Construct the confidence interval

c. Find the length of the confidence interval

Answer 1

1.

a) The critical value for $\alpha = 0.1$ is $z_c = z_{1-\alpha/2} = 1.6449.$

The margin of error is

$$ME=z_c\times \dfrac{\sigma}{\sqrt{n}}=1.6449\times \dfrac{40}{\sqrt{150}}$$

$$\approx5.372221$$

b) The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-ME, \bar{x}+ME)$$

$$\approx(65-5.372221, 65+5.372221)$$

$$=(59.627779, 70.372221)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $59.628 < \mu < 70.372,$ which indicates that we are 90%  confident that the true population mean $\mu$ is contained by the interval $(59.628, 70.372).$

c) The length of the confidence interval is $2ME$

$$2(5.372221)=10.744442$$

2.

a) The critical value for $\alpha = 0.01, df=n-1=349$ degrees of freedom is  $t_c = z_{1-\alpha/2; n-1} = 2.58999$

The margin of error is

$$ME=t_c\times \dfrac{s}{\sqrt{n}}=2.58999\times \dfrac{750}{\sqrt{350}}$$

$$\approx103.8306$$

b) The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-ME, \bar{x}+ME)$$

$$\approx(13500-103.8306, 13500+103.8306)$$

$$=(13396.1694, 13603.8306)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $13396.17 < \mu < 13603.83,$ which indicates that we are 99%  confident that the true population mean $\mu$ is contained by the interval $(13396.17, 13603.83).$

c) The length of the confidence interval is $2ME$

$$2(103.8306)=207.66$$