The CEO of a large cable company claims that 75 percent of his 1,000,000 customers are satisfied with the service they receive. To test this claim, the local newspaper surveyed 100 customers, using simple random sampling. 76 were satisfied and the remaining were unsatisfied. Based on these findings from the sample, can we reject the CEO’s hypothesis that 75% of the customers are satisfied? Use a .05 level of significance. Include the null and alternative hypothesis, test statistic, p-value, decision, and conclusion.

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A: Given: Sample size,n= 76 Number of students who have jobs,x= 35  Level of significance=0.05

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A: Given information: The median rental, M=$1200. From a sample of 149 units, 5 units were rented for…

Q: In the US, 43.5% of all people have type O blood, 39.7% have type A blood, 11.9% have type B blood…

A: Outcomes Frequency O1271A1123B328AB158Total2880

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A:

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A: Given Aplha = 0.05 Df=N-1    =3-1=2

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Q: A union of restaurant and foodservice workers would like to estimate the mean hourly wage, u, of…

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Q: Why is it necessary for a sample to be obtained randomly rather than conveniently?

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Q: Based on these findings, can we reject the CEO's hypothesis that 80% of the customers are very…

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Q: A radio news story claimed that half of all U.S. adults have Internet access. In a national poll…

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Q: A polling organization conducted telephone surveys in March of 2004, 2005 and 2006. In each year,…

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A: Sample Size:The formula for the sample size is as follows:

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A: It is given that sample size (n) is 4,625 and p hat value is 1-0.32 = 0.68.

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