Question

Young's modulus is a quantitative measure of stiffness of an elastic material. Suppose that for aluminum alloy sheets of a particular type, its mean value and standard deviation are 70 GPa and 1.6 GPa, respectively.
a. If X is the sample mean Young's modulus for a random sample of n = 16 sheets, where is the sampling distribution of X centered, and what is the standard deviation of the X distribution?
b. Answer the questions posed in part (a) for a sample size of n = 64
c. For which of the two random samples, the one of part (a) or the one of part (b), is X more likely to be within 1 GPa of 70 GPa? Explain your reasoning.

Answer 1

Answer:

a1) $\mu_{\bar{X}} = 70$

a2)$\sigma_{\bar{X}} = 0.4$

b1) $\mu_{\bar{X}} = 70$

b2) $\sigma_{\bar{X}} = 0.2$

c) X is more likely to be within 1 GPa of 70 GPa in the random sample of part b because of the largeness in sample size and less scattering of data

Step-by-step explanation:

Mean value, $\mu = 70$

Standard deviation, $\sigma = 1.6$

a1) sample size, n = 16

Mean of the sampling distribution of the sample mean = mean value, i.e.

$\mu_{\bar{X}} = \mu\\\mu_{\bar{X}} = 70$

a2) The standard deviation of the sampling distribution of the sample mean

$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n} } \\\sigma_{\bar{X}} = \frac{1.6}{\sqrt{16} }\\\sigma_{\bar{X}} = 0.4$

b1) For sample size, n = 64

Mean of the sampling distribution of the sample mean = mean value, i.e.

$\mu_{\bar{X}} = \mu\\\mu_{\bar{X}} = 70$

a2) The standard deviation of the sampling distribution of the sample mean

$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n} } \\\sigma_{\bar{X}} = \frac{1.6}{\sqrt{64} }\\\sigma_{\bar{X}} = 0.2$

c) X is more likely to be within 1 GPa of 70 GPa in the random sample of part b because it has a larger sample size, hence a decrease in the variability. This makes us easily determine the position of the sample around the population mean