Question

A consumer electronics company is comparing the brightness of two different types of picture tubes for use in its television sets. Tube type A has mean brightness of 100 and standard deviation of 16, and tube type B has unknown mean brightness, but the standard deviation is assumed to be identical to that for type A. A random sample of tubes of each type is selected, and is computed. If equals or exceeds , the manufacturer would like to adopt type B for use. The observed difference is . a. What is the probability that exceeds by 3.0 or more if and are equal

Answer 1

Answer:

The answer is "0.7794".

Step-by-step explanation:

Please find the complete question in the attached file.

Given:

$\to n_{1}=n_{2}=25$

Hypotheses:

$\to H_{0}:\mu_{B}-\mu_{A}\geq 0\\\\\to H_{a}:\mu_{B}-\mu_{A}< 0$

Testing statistics:

$\to z=\frac{(\bar{x}_{B}-\bar{x}_{A})-(\mu_{B}-\mu_{A})}{\sqrt{\frac{\sigma^{2}_{B}}{n_{1}}+\frac{\sigma^{2}_{A}}{n_{2}}}}=\frac{3.5-(0)}{\sqrt{\frac{16^{2}}{25}+\frac{16^{2}}{25}}}=0.77$

The test is done just so the p-value of a test is

$\to p-value = P(z < 0.77) = 0.7794$

Because the p-value of the management is large, type B can take it.