Question

Data on return-to-pay ratios was collected from CEOs of companies within both the low-tech industry and the consumer products industry.
Low-Tech Consumer Products
Sample size 14 12
Sample mean 157 218
Sample Variance 1563 1602
Assume population variances are unequal.

(a) The point estimate of the difference between the means of the two populations is

(b) The standard error for the difference between the two means is

(c) The correct distribution to use is :

t-distribution with 26 degrees of freedom

t-distribution with 23 degrees of freedom

normal distribution

t-distribution with 24 degrees of freedom

Answer 1

Answer:

Step-by-step explanation:

Low-Tech Consumer Products

Sample size 14 12

Sample mean 157 218

Sample Variance 1563 1602

Sample S.d         √1563 √1602

Assume population variances are unequal.

(a) The point estimate of the difference between the means of the two populations is

$=\bar x_1 - \bar x_2\\\\=218-157=61$

(b) The standard error for the difference between the two means is

$SE = \sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2} } \\\\=\sqrt{\frac{1602}{12} +\frac{1563}{14} } \\\\=\sqrt{133.5+111.643} \\\\=\sqrt{245.143} \\\\=15.657$

(c) The correct distribution to use is

$=\frac{\frac{S_1^2}{n_2}+\frac{S_2^2}{n_2} }{(\frac{S_1^2}{n_1})^2 /n_1 -1+(\frac{S_2^2}{n_2})^2 n_2-1 }$

$=\frac{(133.5+111.643)^2}{\frac{(133.5)^2}{11} +\frac{(111.643)^2}{13} } \\\\=\frac{60095.09045}{2578.986041} \\\\=23.30\approx23\ \text{nearest interger}$

t-distribution with 23 degrees of freedom

Answer 2

Answer:

Step-by-step explanation:

The confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

Where

x1 = sample mean of low-tech industry

x2 = sample mean of consumer products industry

s1 = sample standard deviation low-tech industry

s2 = sample standard deviation for consumer products industry

n1 = number of samples of low-tech industry

n2 = number of samples of consumer products industry

a) x1 - x2 is the point estimate of the difference between the means of the two populations

Therefore,

Point estimate = 157 - 218 = - 61

b) the formula for standard error is expressed as

√(s1²/n1 + s2²/n2)

Variance = standard deviation²(s²)

s1² = 1563

s2² = 1602

Standard error = √(1563²/14 + 1602²/12) = 623.2

c) Degree of freedom =

(n1 - 1) + (n2 - 1) = (14 - 1) + (12 - 1) = 24

t-distribution with 24 degrees of freedom