Question

1.

In a random sample of 86 college students has a mean earnings of $3120 with a standard deviation of $677 over the summer months. Determine whether a normal distribution (Z values) or a t- distribution should be used or whether neither of these can be used to construct a confidence interval. Assume the distribution of weekly food expense is normally shaped.

	a.	Use a t distribution
	b.	Use a normal distribution (Z values)
	c.	Neither a normal distribution nor a t distribution can be used

2.

Find the critical t value for a 90% confidence interval and a sample size, n, equal to 10

	a.	2.262
	b.	3.25
	c.	1.812
	d.	1.833
	e.	None of the above

3.

What are the critical Z values for a 95% Confidence Interval

	a.	-1.25 and 1.25
	b.	-1.645 and 1.645
	c.	-1.575 and 1.575
	d.	-1.96 and 1.96
	e.	None of the above

4.

What are the critical Z values for a 90% Confidence Interval

	a.	-1.25 and 1.25
	b.	-1.645 and 1.645
	c.	-1.575 and 1.575
	d.	-1.96 and 1.96
	e.	-2.33 and 2.33

Answer 1

Solution:

1.

Population SD is unknown. So use t distribution.

Use a t distribution

2.

c = 90% = 0.90   

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05

Also, d.f = n - 1 = 10 - 1 = 9

     =  _0.05,9 = 1.833

Answer : 1.833

3.

c = 95% = 0.95   

= 1- c = 1- 0.95 = 0.05

  /2 = 0.025

Using z table , z = 1.96

-1.96 and 1.96

4.

c = 90% = 0.90   

= 1- c = 1- 0.90 = 0.10

  /2 = 0.05

Using z table , z = 1.645

-1.645 and 1.645