Tuesday, April 14, 2009

April 18, 2009 - Homework, Part II

Complete the following homework activity and post as a group:

From your textbook
Elementary Statistics—Picturing the World:

1. Section 6.1, Exercise #24, p. 288

2. Section 6.1, Exercise #34, p. 288

3. Section 6.2, Exercise #8, p. 300

4. Section 6.2, Exercise #12, p. 300
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5 comments:

  1. AlexApril 18, 2009 at 9:40 AM

    Here goes nothing...

    #24: c = 0.95, x=31.39, s=0.8, n=82
    E = 1.96 * 0.8 / Sqrt of 82 = ~1.2
    X- E = 31.39 – 1.2 = 30.19, 31.39 + 1.2 = 32.59
    30.19 < 1.2 < 32.59
    With 95% confidence, I can say that the confidence interval is between 30.19 and 32.59

    #34: Given sample mean and sample standard deviation , construct 90% and 95%. Which is wider?
    Random sample of 36 days , mean is 19.31 standard dev is 2.37
    C = .90, .95, x = 19.31, s = 2.37, n = 36
    E 1= 1.645 * 2.37 / sqrt(36) = ~0.6
    19.31 – 0.6 = 18.71 , 19.31 + 0.6 = 19.91
    18.71 < 0.6 < 19.91
    E2 = 1.96 * 2.37 / sqrt(36) = ~0.78
    19.31 – 0.78 = 18.53 , 19.31 + 0.78 = 20.09
    18.53 < 0.78 < 20.09
    E2, or the 95% confidence interval appears to be wider

    #8: use T-distribution. Using normal distribution, which is wider?
    c=0.95, x=13.4, s=0.85, n=8
    8-1 = 7 or +-2.365 is the critical value of T
    E = 2.365 * 0.85 / sqrt(8) = ~0.71
    13.4 – 0.71 and 13.4 + 0.71 = 12.65, 14.11 or
    12.65 < 0.71 < 14.11
    On a normal distribution:
    E = 1.96 * 0.85 / sqrt(8) = ~0.59
    13.4 – 0.59 and 13.4 + 0.59 = 12.81, 13.99
    12.81 < 0.59 < 13.99
    The T-distribution interval appears to be wider

    #12: 7 computers, mean is 100.00, standard dev is 42.50
    C=.95, x=100.00, s=42.50, n=7
    7-1 = 6, or 2.447
    E= 2.447 * 42.50 / sqrt(7) = ~39.30
    100 – 39.30 and 100 + 39.30 = 60.7 and 139.30
    60.7 < 39.30 < 139.30
    With 95% confidence, I can say that the confidence interval is between 60.7 and 139.30

    ReplyDelete
  2. AlexApril 18, 2009 at 10:43 AM

    Oops, our group was:

    Jace, Justin, Myself

    ReplyDelete
  3. BryanApril 18, 2009 at 11:33 AM

    Posted By: Bryan S., Sheila W., Karolyn M., Nick N.

    24) Then confidence interaval is between 31.23 and 31.55.

    34) 95% confidence level has a wider interval level.

    8) a. The t distrobution is 2.365
    b. The t distrobution seems to be wider.

    12) The confidence level is between 139.24 and 60.76

    ReplyDelete
  4. rickApril 18, 2009 at 12:23 PM

    Evan Dossey, Rick Robinson, Mel Bailey

    1. Section 6.1, Exercise #24, p. 288
    31.21 to 31.56

    2. Section 6.1, Exercise #34, p. 288
    90% - 18.66 to 19.96
    95% - 18.56 to 20.08
    The 95% is wider

    3. Section 6.2, Exercise #8, p. 300
    12.69 to 14.11
    t-distribution would be wider

    4. Section 6.2, Exercise #12, p. 300
    confidence i = 60.756 to 139.24
    margin of error = 39.244

    ReplyDelete
  5. AnonymousApril 18, 2009 at 12:36 PM

    Tonya
    Rick Oneil

    #24. - .173

    #34 - 90% = .651, 95% = .774 The wider interval is 95%.

    #8 - A. = .712, B. = .590 which makes the T-Distribution wider.

    #12– T-Distribution is 2.447, Margin of error E = 39.318
    Confidence intervals = Upper = 139.318, Lower = 60.682

    ReplyDelete

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