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    Confidence Interval and Statistical Analysis

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    Find the confidence interval and conduct statistical analysis for the given problems.

    6. A producer of a juice drink advertises that it contains 10% real fruit juices. A sample of 75 bottles of the drink is analyzed and the percent of real fruit juices is found to be 6.5%. If the true proportion is actually .10, what is the probability that the sample percent will be 6.5% or less? Based on the sample, what would you conclude about the juice producers' claims of "...contains 10% real fruit juices..."?

    7. The security department of a state university is planning its budget for the next year. In estimating the man-hours for security during university sponsored music concerts, the average length of music concerts is needed. A random sample for 36 security departments at universities was taken and the sample mean length of concerts was 160 minutes. Suppose the standard deviation is 45 minutes. Prepare an 84 percent confidence interval for the true mean duration of music concerts.

    8. A bank holds the mortgage on a motel that the motel owner says has an average daily revenue of $2048. In the event the average is not at least $2048, the bank will ask the motel owner for additional collateral for the loan as protection against loan default. A loan officer is aware that the daily average revenue is normally distributed and randomly selects 20 days during the last six months. The sample has a mean of $2003 and standard deviation of $144. Based on an appropriate statistical analysis and the sample taken by the loan office, determine whether the bank should ask for additional collateral.

    9. The tourist bureau of a large populous state is concerned about the motel space available for holding a world's fair. Previous occupancy rates have been 60%. In order to determine whether any change in the occupancy rate has occurred the bureau takes a random survey of 500 motels in the state. The reported occupancy rate of these 500 motels is 65%. Based on the sample taken, and an appropriate statistical analysis, determine the maximum level of confidence at which it can be said that the occupancy rates are not as believed to be previously.

    10. A long-haul truck drive is eligible for a year-end bonus if the average miles driven per day are more than 375 miles per eight-hour day. To determine whether or not a drive is to receive a year-end bonus, a random sample of 36 days from the driver's logbook is taken and the sample mean of 405 miles and standard deviation of 70 is computed. What is the maximum level of confidence at which it can be said that this driver is eligible for a bonus?

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    Solution Preview

    p= 0.1
    n =75
    => m = mean = n*p =75*0.1 = 7.5
    P[X<= 6.5] = 0.45 (from Poisson table) --Answer
    It means one can conclude that ~55% there are chances that the producer's claim is correct.

    n = 36,
    ample mean = <X> = 160 min
    s (standard deviation ) = 45 min
    true mean = m =?
    confidence level =84% = (1-a)*100
    => a = 0.16
    => a/2 = 0.08
    P[<X> -z(a/2)*s/sqrt(n) <= m <= <X> + z(a/2)*s/sqrt(n)] = 0.84
    because z(a/2) = z(0.08) = 1.405
    => ...

    Solution Summary

    The solution answers the question(s) below. The confidence intervals and statistical analysis is examined.