Question

marketing firm asked a random set of married and single men how much they were willing to spend on a vacation. At α = 0.05, we wish to test whether there is a difference in the two amounts. What is the P-value of this test? Round your answer to 3 significant places after the decimal. (For example, 0.0678 has to be entered as 0.068) Married men Single men Sample size 60 60 Sample mean $880 $845 Population variance 5200 7100

Answer 1

Answer:

The P-value is 0.016.

Step-by-step explanation:

Given information:

Significance level, α = 0.05

                            Married men         Single men

Sample size               60                            60

Sample mean          $880                        $845

Population variance  5200                       7100

We need to test whether there is a difference in the two amounts.

Null hypothesis: $H_0:\mu_1=\mu_2$

Alternative hypothesis: $H_1:\mu_1\neq \mu_2$

The formula for t-statistics is

$t=\frac{(\overline{x}_1-\overline{x}_2)+(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}$

$t=\frac{(880-845)+(0)}{\sqrt{\frac{5200}{60}+\frac{7100}{60}}}$

$t=2.44450603519$

$t\approx 2.44$

Degree of freedom is

$d.f=n_1+n_2-2=60+60-2=118$

The P-value for $t=2.44$ at significance level α = 0.05, wih degree of freedom 119 is 0.016174. Approximately it can be written as 0.016.

Therefore the P-value is 0.016.