Applications The Employment and Training Administration reco

Applications The Employment and Training Administration recorted that the U.S. mean unemploymeasts nce benefit was $238 per week (The World Almanac, 2003). A insurance of Virginia anticipated that sample data would show evidence that the ean employment insurance benefit in Virginia was below the national average. Almanac, 2003). A researcher in the state hat the mean weekly un- Develop appropriate hypothesessuch that rejection of 11° will support the rese . a contention. b. For a sample of 100 individuals, the sample mean weekly unemployime benefit was $231 with a sample standard deviation of $80. What is At = .05, what is your conclusion? Repeat the preceding hypothesis test using the critical value a sample mean weekly unemployment insurance the p-value? d. pproach. 28. A shareholders\' group, in lodging a protest, claimed that the mean tenure for a chict eee a chief exec- tive office (CEO) was at least nine years. A lournal found a sample mean ten of s 6.38 years (The Wall Street Journal, January 2, 2007). a. Formulate hypotheses that can be used to challenge the validity of the claim made by al found a sample mean tenure of survey of companies reported in The Wall Street ure of a 7.27 years for CEOs with a standard deviation 7.27 years for CEOs with a s the shareholders\' group b. Assume 85 companies were included in the sample. What is the p-value for your hy- h. Asume s comsere included mhat is the pvalie for y pothesis test? C. At = .01, what is your conclusion? The cost of a one-carat VS2 clarity, H color diamond from Diamond Source USA is $5600 Diamond Source website, March 2003). A midwestern jeweler makes calls to contacts in he diamond district of New York City to see whether the mean price of diamonds there dif- rs from $5600. 2 Formulate hypotheses that can be used to determine whether the mean price in New York City differs from $5600.

Solution

QNo27

H0: xbar=mu

Ha: xbar <mu

(one tailed test)

3

11

Sample Mean:

Population Mean :

Sample Standard Deviation:

Population Size:

Significance Level ():

0.5

T statistic =

-1.23744

P value =

0.1080

p>0.05

reject H0

Researchers claim is correct

28)

H0: xbar = mu

Ha: xbar >9

Sample mean = 7.27

N = 85

S=6.38

T statistic =

-2.4997

P value=

0.9938

Since the P-value is > than the significance level 0.01, we cannot reject the null hypothesis.

Mu = 9

Conclusion: sample mean cannot be greater than 9

QNo27 H0: xbar=mu Ha: xbar <mu (one tailed test)							3
							11
Sample Mean: Population Mean : Sample Standard Deviation: Population Size: Significance Level (): 0.5 T statistic = -1.23744 P value = 0.1080 p>0.05 reject H0 Researchers claim is correct 28) H0: xbar = mu Ha: xbar >9 Sample mean = 7.27 N = 85 S=6.38 T statistic = -2.4997 P value= 0.9938 Since the P-value is > than the significance level 0.01, we cannot reject the null hypothesis. Mu = 9