Question

A sample of 56 research cotton samples resulted in a sample average percentage elongation of 8.17 and a sample standard deviation of 1.42 ("An Apparent Relation Between the Spiral Angle f, the Percent Elongation E1, and the Dimensions of the Cotton Fiber," Textile Research J., 1978: 407- 410). Calculate a 95% large-sample CI for the true average percentage elongation ?. What assumptions are you making about the distribution of percentage elongation?

Answer 1

Answer:

(7.7981, 8.5419). The distribution of percentage elongation can be any distribution with finite mean and variance.

Step-by-step explanation:

We have a large sample size n = 56 research cotton samples. Besides, $\bar{x} = 8.17$ and $s = 1.42$. A 95% large-sample CI for the true average percentage elongation is given by $\bar{x}\pm z_{0.05/2}(s/\sqrt{n})$, i.e., $8.17\pm (1.96)(1.42/\sqrt{56})$ or equivalently (7.7981, 8.5419). The distribution of percentage elongation can be any distribution with finite mean and variance because we have a large sample.