Question

QUESTION: A) In week 1 you calculated the mean number of hours on electronic devices for the 10 people in your personal sample. Use this value as the starting point to create a 90% confidence interval for the number of hours a day on electronic devices a person has. The actual average of our class data is 6.2 hours (with standard deviation of 3.4). Does your confidence interval contain the actual value? Discuss your results with your reasoning as to why you think you obtained the results you did?

B) Using the scenario from question 2, how many people would need to be selected to be 90% confident that the sample mean number of hours would be within ¼ hour of the actual mean?

CLASS DATA:

Mean number of hours on electronic device from sample of 10 people = 6.5 hours

Standard Dev.: 2.64

Answer 1

Solution:

A)

Given:

The sample mean is =6.2 and the sample standard deviation is s = 3.4

The size of the sample is n = 10

The confidence level is 90% = 0.90

Therefore, level of significance = 1-0.90 = 0.10

The critical t-value for α=0.1 and df = 10-2 = 9 is,

tc​=1.833 ...Using excel function, =TINV(0.1,9)

The 90% confidence for the population mean μ:

Calculation:

= (6.2−1.971 , 6.2+1.971)

= (4.229 , 8.171)

The actual value is 10. This value does not contains in the above confidence interval. As the sample mean is very lower as compare to population mean and sample size also less which gives insufficine result.

B)

Given:

Confidence level = 0.90

Level of significance = 1-0.90

Margin of error = E = 1/4 = 0.25

Standard deviation = σ=2.64

Therefore, Z_c = Z_α/2 = 1.64 ..Using excel formula =ABS(NORMSINV(0.05)

The sample size can be calculated as,

n 302

Therefore , number of people to be select = 302

Done