Question

You are babysitting children when you notice that one of them may have a fever. You check her temperature four times with the following results: 101 101 97 101

Compute the sample mean (x̄) Select one: a. 100 b. 101 c. 200 d. 97

Compute the sample standard deviation (s) Select one: a. 1 b. 2 c. 1.73 d. 0.86

Compute the standard error (SE) Select one: a. 1 b. 2 c. 1.73 d. 0.86

What t* would you use for a 90% confidence interval? Select one: a. 2.35 b. 2.13 c. 1.64 d. 1.53

What t* would you use for a 95% confidence interval? Select one: a. 3.18 b. 2.78 c. 2.13 d. 2.35

What is the 90% confidence interval for the baby's temperature? Select one: a. (97.65, 102.35) b. (97.87, 102.13) c. (98.36, 101.64) d. (98.47, 101.53)

What is the 95% confidence interval for the baby's temperature? Select one: a. (96.82, 103.18) b. (97.22, 102.78) c. (97.87, 102.13) d. (97.65, 102.35)

Answer 1

Answer:

Given,

101 , 101 , 97 , 101

Sample mean = x/n

= (101 + 101 + 97 + 101) / 4

= 100

Standard deviation = sqrt((x-xbar)^2/(n-1))

substitute values

= sqrt((101-100)^2 + (101-100)^2 + (97-100)^2 + (101-100)^2)/(4-1))

= 2

Standard error = s/sqrt(n)

= 2/sqrt(4)

= 1

For 90% CI, alpha = 0.10

t = t(alpha/2 , df) = t(0.10/2 , 3) = 2.35

95% CI , alpha = 0.05

t = t(alpha/2 ,df) = t(0.05/2 , 3) = 3.18

90% CI = xbar +/- t*s

= 100 +/- 2.35*1

= (97.65 , 102.35)

95% CI = xbar +/- t*s

= 100 +/- 3.18*1

= (96.82 , 103.18)