Question

1.)

You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is σ=49.3σ=49.3. You would like to be 99% confident that you estimate is within 5 of the true population mean.

How large of a sample size is required?

n =  

Note: Use z-scores rounded to 3 decimal places in your calculations.

2.)

You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately σ=26.1σ=26.1. You would like to be 99% confident that you estimate is within 3 of the true population mean. How large of a sample size is required?

n =



Note: Use z-scores rounded to 3 decimal places.

3.)

Internet and Mobile Devices. In a 2012 survey of approximately 2275 American adults, 387 of them do their browsing on their phone rather than a computer or other device.

(a) Calculate a 98% confidence interval for the proportion of Americans who access the internet on their cell phones, and interpret the interval in this context.

(  ,  )

(b) Interpret the confidence interval in the context of the study:

• We can be 98% confident that the proportion of Americans who use their cell phones to access the internet is in this confidence interval
• 98% of Americans use their cell phone to access the internet
• We can be 98% confident that our confidence interval contains the sample proportion of Americans who use their cell phones to access the internet

Answer 1

1)

Solution :

Given that,

Population standard deviation = = 49.3

Margin of error = E = 5

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

sample size = n = (Z_/2* / E) ²

n = (2.576 * 49. 3/ 5)²

n = 645.12

n = 646

Sample size = 646