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

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

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/√(n))`, i.e.,
`8.17\pm (1.96)(1.42/√(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.