Answer:
The answer is "0.7794".
Step-by-step explanation:
Please find the complete question in the attached file.
Given:
[tex]\to n_{1}=n_{2}=25\\\\[/tex]
Hypotheses:
[tex]\to H_{0}:\mu_{B}-\mu_{A}\geq 0\\\\\to H_{a}:\mu_{B}-\mu_{A}< 0\\\\[/tex]
Testing statistics:
[tex]\to z=\frac{(\bar{x}_{B}-\bar{x}_{A})-(\mu_{B}-\mu_{A})}{\sqrt{\frac{\sigma^{2}_{B}}{n_{1}}+\frac{\sigma^{2}_{A}}{n_{2}}}}=\frac{3.5-(0)}{\sqrt{\frac{16^{2}}{25}+\frac{16^{2}}{25}}}=0.77[/tex]
The test is done just so the p-value of a test is
[tex]\to p-value = P(z < 0.77) = 0.7794[/tex]
Because the p-value of the management is large, type B can take it.
