Question

Assume that a sample is used to estimate a population mean μμ. Find the 99.5% confidence interval for a sample of size 642 with a mean of 22.2 and a standard deviation of 9.3. Enter your answers accurate to four decimal places.

Confidence Interval = ( , )

You measure 29 textbooks' weights, and find they have a mean weight of 47 ounces. Assume the population standard deviation is 10.8 ounces. Based on this, construct a 99.5% confidence interval for the true population mean textbook weight.

Keep 4 decimal places of accuracy in any calculations you do. Report your answers to four decimal places.

Confidence Interval = ( , )

Answer 1

Solution :

Given that,

1)Point estimate = sample mean = = 22.2

Population standard deviation = = 9.3

Sample size = n = 642

At 99.5% confidence level

= 1-0.995% =1-0.995 =0.005

/2 =0.005/ 2= 0.0025

Z/2 = Z_{0.0025 = 2.807}

Z/2 = 2.807
Margin of error = E = Z/2 * ( /n)

= 2.807 * ( 9.3 /  642 )

= 1.0303

At 99.5 % confidence interval estimate of the population mean is,

- E < < + E

22.2 - 1.0303 <   < 22.2 + 1.0303

21.1697 <   < 23.2303

Confidence Interval( 21.1697 ,23.2303 )

2)

Point estimate = sample mean = = 47

Population standard deviation = =10.8

Sample size = n = 29

At 99.5% confidence level

= 1-0.995% =1-0.995 =0.005

/2 =0.005/ 2= 0.0025

Z/2 = Z_{0.0025 = 2.807}

Z/2 = 2.807
Margin of error = E = Z/2 * ( /n)

= 2.807 * ( 10.8 /  29 )
=5.6295

At 99.5 % confidence interval estimate of the population mean is,

- E < < + E
47 - 5.6295 <   < 47 + 5.6295
41.3705 <   < 52.6295

Confidence Interval ( 21.1697 ,52.6295 )