Question

A shareholders' group is lodging a protest against your company. The shareholders group claimed that the mean tenure for a chief exective office (CEO) was at least 8 years. A survey of 79 companies reported in The Wall Street Journal found a sample mean tenure of 6.6 years for CEOs with a standard deviation of 5.5 years (The Wall Street Journal, January 2, 2007).

A shareholders' group is lodging a protest against your company. The shareholders group cdaimed that the mean tenure for a chief exective office (CEO) wasa Questions least 8 years. A survey of 79 companies reported in The Wall Q1 (0/1) Q2 (0/1) Q3 (0/3) Q4 (0/3) Q5 (0/5) Sreet Journal founda sample mean tenure of 6.6 years for CEOS with a standard deviation of 5.5 years (The Wall Street Journal, January 2, 2007) You want to formuate andtest a hypothesis that can be used to challenge the validty of the caim made by the group, at a significance level of 0.10. Your hypotheses are: Q6 (0/5) Ho:u 8 Q7 (0/2) Q8 (0/1) Q9 (0/2) Q10 (0/1) Q11 (0/4) What is the test statistic for this sample? (Report answer accurate to 3 decimal places.) What is the p-value for this sample? Q12 (0/4) Q13 (0/4) (Report answer accurate to 4 decimal places.) Q14 (0/4) Q15 (0/4) The p-value is O less than (or equal to) a Q16 (0/4) Q17 (0/4) O greater than o Q18 (0/4) Q19 (0/6) Q20 (0/4) This test statistic leads to a decision to Q21 (0/2) Oreject the nul Grade: 0/68 O accept the null Print Version O fail to reject the null As such, the final conclusion is that O there is sufficient evidence to conclude that the popuation mean tenure for CEOS is equal to 8 Othere is not sufficient evidence to concdlude that the population mean tenure for CEOS is equal to 8 Othere is sufficient evidence to conclude that the population mean tenure for CEOs is less than 8 Othere is not sufficient evidence to condude that the population mean tenure for CEOs is less than 8.

Answer 1

Solution :

Given that,

Population mean = $\mu$ = 7

Sample mean = $\bar x$ = 6.6

Sample standard deviation = s = 5.5

Sample size = n = 79

Level of significance = $\alpha$ = 0.10

This is a left tailed test.

The test statistics,

t = ( $\bar x$ - $\mu$ )/ (s/$\sqrt{n}$)

= ( 6.6 - 7 ) / ( 5.5 / $\sqrt{}$ 79 )

= -0.539

P- Value = 0.2958

Greater than $\alpha$ .

Fail to reject null hypothesis.

There is not sufficient evidence to conclude that the population mean tenure for CEOs is less than 8.