Answer to Question #347847 in Statistics and Probability for eros

Question

1.    Assuming that the samples come from normal distributions, find the margin of error  given the following:

a. n = 10 and X = 28 with s = 4.0, 90% confidence

b. n = 16 and X = 50 with s = 4.2, 95% confidence

c. n = 20 and X = 68.2 with s = 2.5, 90% confidence

d. n = 23 and X = 80.6 with s = 3.2, 95% confidence

e. n = 25 and X = 92.8 with s = 2.6, 99% confidence

 

2.    Using the information in number 2, find the interval estimates of the population mean.

Answer 1

1.

a.

The critical value for $\alpha = 0.10, df=n-1=9$ degrees of freedom is $t_c​=z_{1−α/2;n−1}=1.833113.$

$$E=t_c\times\dfrac{s}{\sqrt{n}}=1.833113\times\dfrac{4}{\sqrt{10}}=2.3187$$

b.

The critical value for $\alpha = 0.05, df=n-1=15$ degrees of freedom is $t_c​=z_{1−α/2;n−1}=2.131449.$

$$E=t_c\times\dfrac{s}{\sqrt{n}}=2.131449\times\dfrac{4.2}{\sqrt{16}}=2.2380$$

c.

The critical value for $\alpha = 0.10, df=n-1=19$ degrees of freedom is $t_c​=z_{1−α/2;n−1}=1.729133.$

$$E=t_c\times\dfrac{s}{\sqrt{n}}=1.729133\times\dfrac{2.5}{\sqrt{20}}=0.9666$$

d.

The critical value for $\alpha = 0.05, df=n-1=22$ degrees of freedom is $t_c​=z_{1−α/2;n−1}=2.073873.$

$$E=t_c\times\dfrac{s}{\sqrt{n}}=2.073873\times\dfrac{3.2}{\sqrt{23}}=1.3838$$

e.

The critical value for $\alpha = 0.01, df=n-1=24$ degrees of freedom is $t_c​=z_{1−α/2;n−1}=2.79694.$

$$E=t_c\times\dfrac{s}{\sqrt{n}}=2.79694\times\dfrac{2.6}{\sqrt{25}}=1.4544$$

2.

a.

$$\mu=28\pm2.3187$$

$$(25.6813, 30.3187)$$

b.

$$\mu=50\pm2.2380$$

$$(47.7620, 52.2380)$$

c.

$$\mu=68.2\pm0.9666$$

$$(67.2334, 69.1666)$$

d.

$$\mu=80.6\pm1.3838$$

$$(79.2162, 81.9838)$$

e.

$$\mu=92.8\pm1.4544$$

$$(91.3456, 94.2544)$$