Question

The diameter of small Nerf balls manufactured at a factory in China is expected to be approximately normally distributed with a mean of 5.2 inches and a standard deviation of .08 inches. Suppose a random sample of 20 balls is selected. Find the interval that contains 95.44 percent of the sample means.

[5.07, 5.33]

[5.16, 5.24]

[5.04, 5.36]

[5.18, 5.22]

Answer 1

Solution :

Given that,

= 5.2

s =0.08

n =20

Degrees of freedom = df = n - 1 =20 - 1 = 19

a ) At 95.44% confidence level the t is ,

= 1 - 95.44% = 1 - 0.9544 = 0.0456

  / 2= 0.0456 / 2 = 0.0228

t _/2,df = t_0.0228,19 = 2.139 ( using student t table)

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

=2.139 * (0.08 / 20)

=0.0383

The 95.44% confidence interval estimate of the population mean is,

- E < < + E

5.2- 0.0383< < 5.2+ 0.0383

5.16 < < 5.24

( 5.16 , 5.24)