Question

Question 7 (4.2 points) A simple random sample of electronic components will be selected to test for the mean lifetime in hours. Assume that component lifetimes are normally distributed with population standard deviation of 20 hours. How many components must be sampled so that a 99% confidence interval will have margin of error of 6 hours? Write only an integer as your answer. Question 8 (5 points) Six measurements were made of the mineral content (in percent) of spinach, with the following results. It is reasonable to assume that the population is approximately normal. 19.1, 20.1, 20.8, 21.6 , 20.5, 19.1 Find the lower bound of the 95% confidence interval for the true mineral content. Round to three decimal places (for example: 20.015). Write only a number as your answer. Do not write any units.

Answer 1

Answer:

7)

c= 99% , $\sigma$ =20, E=6

formula for sample size is

Zc*02 n = (

Where Zc is the z critical value for the c= 99%

Zc = 2.575

2.575 * 202 n =-

n = 73.67361111

n= 74

Answer = 74

8)

n=6, C=95%

now calculate the sample mean and sample standard deviation for

the given sample data we get,

$\bar x$ = 20.2, s =0.98387

formula for confidence interval is

Itc*

Where tc is the t critical value for c= 95%, with df = n-1 =-1 =5

tc = 2.571

10.98387 20.2±2.571 + = v6

19.16732 < $\mu$ < 21.23897

19.167 < $\mu$ < 21.233

Lower Bound = 19.167