Question

A local company makes a candy that is supposed to weigh 1.00 ounces. A random sample of 25 pieces of candy produces a mean of 0.996 ounces with a standard deviation of 0.004 ounces. Construct a 98 percent confidence interval for the mean weight of all such candy. Assume that X, the weight of a candy, is distributed normally.

Answer 1

Answer: (0.994, 0.998)

Step-by-step explanation:

Given : Sample size : n= 25 < 30 , so we use t-test.

$\overline{x}=0.996\ \ ;\ s=0.004$

Critical t-value use for 98% confidence :

$t_{n-1,\alpha/2}=t_{24,0.01}=2.492$

Formula to find the confidence interval :-

$\overline{x}\pm t_{\alpha/2}(\dfrac{\sigma}{\sqrt{n}})$

i.e. $0.996\pm (2.492)(\dfrac{0.004}{\sqrt{25}})$

$=0.996\pm 0.0019936\approx0.996\pm0.002\\\\=(0.996-0.002,\ 0.996+0.002)\\\\=(0.994,\ 0.998)$

Hence, the required confidence interval : (0.994, 0.998)