Question

Assume that we want to construct a confidence interval. Do one of the​ following, as​ appropriate: (a) find the critical value t alpha/ 2​, ​(b) find the critical value z Subscript alpha/02​, or​ (c) state that neither the normal distribution nor the t distribution applies. Here are summary statistics for randomly selected weights of newborn​ girls: n=291​, x over b+31.3 ​hg, s=7.6 hg. The confidence level is 90​%.

Answer 1

Solution :

Given that,

= 31.3

s = 7.6

n = 291

Degrees of freedom = df = n - 1 = 291 - 1 = 290

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,290 = 1.650

The critical value t = 1.650

Margin of error = E = t_/2,df * (s /n)

= 1.650* (7.6 / 291)

= 0.73

Margin of error = 0.73

The 90% confidence interval estimate of the population mean is,

- E < < + E

31.3 - 0.73 < < 31.3 + 0.73

30.56 < < 32.03

(30.56, 32.03 )