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
Accepted Solution
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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 $845Population variance 5200 7100We need to test whether there is a difference in the two amounts.Null hypothesis: [tex]H_0:\mu_1=\mu_2[/tex]Alternative hypothesis: [tex]H_1:\mu_1\neq \mu_2[/tex]The formula for t-statistics is[tex]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}}}[/tex][tex]t=\frac{(880-845)+(0)}{\sqrt{\frac{5200}{60}+\frac{7100}{60}}}[/tex][tex]t=2.44450603519[/tex][tex]t\approx 2.44[/tex]Degree of freedom is[tex]d.f=n_1+n_2-2=60+60-2=118[/tex]The P-value for [tex]t=2.44[/tex] 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.