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Thread: Need some helps with my homework please...

  1. #1
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    Need some helps with my homework please...




    first i want to say hi, my name is jose, and what a cool lil site you guys got here!!!

    im having some issues with my stats hw, there are like 12 questions i cant seem to figure out and/or where to start...maybe you guys can help me out...

    here are my first 2:
    1) According to an irs study, it takes an average of 330 minutes for taxpayers to prepare copy and elec. file 1040 tax form. Assuming a population deviation of 80 min. a consumer watchdog agency selects a random sample of 40 tax payers and conducts a testing. Determine the likelihood that the sample mean is...
    a)greater than 320 minutes?
    b)between 320 and 350 minutes?
    c)greater then 350 minutes?

    2) Dr. Warwick is a prof of English. Recently she counted the number of words mispelled in a group of student essays and found that it follows a standard of deviation of 2.44 words per essay. For the evening class of a sample of 40 students, the mean number of misspelled words was 6.05.
    a)What distribution would you use for the above problem?
    b)Develop a 90% confidence interval for the population mean?
    c)Develop a 95% confidence interval for the population mean?
    d)Develop a 99% confidence interval for the population mean?


    any help is appreciated...

  2. #2
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    Quote Originally Posted by Kidcaffeine View Post
    first i want to say hi, my name is jose, and what a cool lil site you guys got here!!!

    im having some issues with my stats hw, there are like 12 questions i cant seem to figure out and/or where to start...maybe you guys can help me out...

    here are my first 2:
    Sup Jose, thanks for the warning that these are the first of many.

    Quote Originally Posted by Kidcaffeine View Post
    1) According to an irs study, it takes an average of 330 minutes for taxpayers to prepare copy and elec. file 1040 tax form. Assuming a population deviation of 80 min. a consumer watchdog agency selects a random sample of 40 tax payers and conducts a testing. Determine the likelihood that the sample mean is...
    a)greater than 320 minutes?
    b)between 320 and 350 minutes?
    c)greater then 350 minutes?
    First, you need to know the distribution of the sample mean. If your lecturer/class text are any good, you should be able to look this up. The answer is that the sample mean is normally distributed, mean=330 (i.e. same as the population mean, found by the IRS study), and stdev=80/sqrt(40) (NOT the same as the population stdev)

    a) You find P[X > 320] = P[Z > (320-330) / (80/sqrt(40))] = 1 - Phi(-0.79) = Phi(0.79) = .7852

    If you don't know how that Z and Phi got involved, you're in trouble with this problem.

    b) Find P[ 320 < X < 350 ]
    c) Find P[X > 350]

    Quote Originally Posted by Kidcaffeine View Post

    2) Dr. Warwick is a prof of English. Recently she counted the number of words mispelled in a group of student essays and found that it follows a standard of deviation of 2.44 words per essay. For the evening class of a sample of 40 students, the mean number of misspelled words was 6.05.
    a)What distribution would you use for the above problem?
    b)Develop a 90% confidence interval for the population mean?
    c)Develop a 95% confidence interval for the population mean?
    d)Develop a 99% confidence interval for the population mean?
    a) "the above problem" is concerned with the number of mispelled words in a student essay. Since the number of mispelled words is either 0, 1, 2, or some whole discrete number, this should narrow down the choices for distribution. It turns out, that a lot of counts of number of errors/workplace accidents follow the Poisson distribution.

    b,c and d) I'm not sure if that sample stdev=2.44 is supposed to play a part in the calculations...I'm not sure how to reconcile it with the sample mean=6.05. Note that the stdev of a Poisson distribution is equal to the square root of the mean (and 2.44 is nearly the sqrt(6.05)).

    If I had to write an answer down, I'd say sample mean = 6.05, sample stdev = sqrt(6.05) = 2.46, then use the formula for confidence interval:
    (sample mean) +/- z[alpha] * (sample stdev) / sqrt(40)

  3. #3
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    wow mean joe, thank you so much...that helped me alot.

    I really appriciate that.

    I'll try the other ones on my own n see where i go with it.

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