Question

My Cengage Digital Learning Social condus investors screen out stocks of alcohol and tobacce makers, firms with poor environmental records, and companies with poor labor practices. Some examples of good," socially conscious companies are Johnson and Johnson Dell Computers, Bank of America, and Home Depot. The question is, are suth stocks overpriced? One measure of value is the P/E, or price-to-earnings ratio High Peratos may indicate a stock is overpriced. For the SAP Stock Index of major stocks, the mean PVE ratio is 19.4. A random sample of 36'ocwly conscious Mocks gave a Pratio sample mean of 1.6 with sample standard deviations 5.4. Does this indicate that the mean P/E ratio of a socially conscious stocks is different either way from the mean P/E ratio of the S&P Stock Index? Use a 0.05. USE SALT (a) What is the level of cance? State the land terrate hypotheses Mol 19.4; H119.4 MH 1941 194 My #194; M 194 () Wat samping wow you use pain the rationale for your choice of samping itution The student's in the same stage and is unknown The standardona, since the same suis farge and is unknown The student's in the sample is farge and is now The standard romance the sample seis yeande is known What is the value of the sample test statistic (Round your answer to three decimal places) (c) Estimate the value Pa 0.250 0.100 Palve < 0.250 -0.050 < P-value < 0.100 0.010 < P-value < 0.050 Palue < 0.010

Answer 1

We frame the Hypothesis

HOJ = 19.4

$H_{1}: \mu \neq 19.4$

To test the
он
we use t - Statistic; because the Population Standard Deviation $'\sigma '$ is unknown.

$t = \frac{\left | \bar{x} - \mu \right |}{S.E(\bar{x})}$

Where

Population Mean = $\mu$ = 19.4

Sample Size = 36

Sample Mean = = 17.6

T

Sample S.D = = 5.4

S

Therefore

$S.E(\bar{x}) = \frac{s}{\sqrt{n}} = \frac{5.4}{\sqrt{36}} = \frac{5.4}{6} = 0.9$

Therefore

$t = \frac{\left | \bar{x} - \mu \right |}{S.E(\bar{x})}$

$\Rightarrow t = \frac{\left | 17.6 - 19.4\right |}{0.9}$

$\Rightarrow t = \frac{\left | -1.8\right |}{0.9}$

$\Rightarrow t = {\left | -2\right |}$

$\therefore \, \, \, {\left | t_{cal}\right |} = 2$

Given Level of Significance = 5% = 0.05

p - Value:

The p-value is 0.053308

Therefore Option (b) is apt..