Question

Assume the weights of 10 year old children are normally distributed with a mean of 90 pounds and a variance of 36.

a. What is the standard deviation of the sampling mean for a sample of size 100?

b. What is the 95% confidence interval for this distribution?

c. What is the 99% confidence interval for this distribution?

A random sample of 225 statistics tutoring sessions was selected and the number of students absent at each session was recorded. The mean was 11.6 and the standard deviation was 4.1. Estimate the number of absences per session. (Remember: What CL do we default to when not specified?)

Answer 1

(1)

(a)

= Population Mean = 90

= Population SD =

n = Sample Size = 100

The standard deviation of the sampling mean for a sample of size 100 = SE = / = 6/ = 0.6

So,

Answer is:

0.6

(b)

= 0.05

From Table, critical values of Z = 1.96

95 % Confidence Interval:

90 (1.96 X 0.6)

= 90 1.176

= ( 88.824 ,91.176)

95 % Confidence Interval:

88.824 < <   91.176

(c)

= 0.01

From Table, critical values of Z = 2.576

95 % Confidence Interval:

90 (2.576 X 0.6)

= 90 1.5456

= ( 88.4544 ,91.5456)

99 % Confidence Interval:

88.4544 <   <    91.5456

(2)

n = Sample Size = 225

= Sample Mean = 11.6

s = Sample SD = 4.1

SE = s/

= 4.1/

= 0.2733

Since is not specified, by default, Take = 0.05

From Table, critical values of Z = 1.96

95 % Confidence Interval:

11.6 (1.96 X 0.2733)

= 11.6 0.5357

= ( 11.0643 ,12.1357)

Estimation of the number of absences per session

11.0643 <   <    12.1357