Back to: Chapter 10: Introduction to Estimation

7 thoughts on “Confidence Interval Width vs. Sample Size, Standard Deviation, and Mean (Q2e–i) (Preview)

  1. Hi jason,

    On part G, the standard error you got was 2/ the sq root of 100 when was 0.2. Then you multiplied by 1.96 you got 0.0392 – I just calculated it, and got 0.392. I would get no marks if i was slightly off?

    Thanks!!

    1. My apologies, I just finished the rest of the video, and noticed that you corrected the error. Sorry Jason!!

      1. Hey – No worries! Good luck on the test!

  2. Hey Jason, I noticed that in e) you calculated 1.96(1.5/square root of 300) all in one go and got 6.5+- 0.1697. However, I generally calculate within the brackets first then * 1.96, resulting in 6.5+- 0.1699. This in effect changes my calculations by a minor difference, but I wonder how this would be received on the exam?

    1. Something more than just rounding is happening with your calculations. Let me repeat the solution one step at a time, and I will round (to four decimals) at each step:

      6.5 ± 1.96 ( 1.5 / √300)

      6.5 ± 1.96 ( 1.5 / 17.3205)

      6.5 ± 1.96 ( 0.0866)

      6.5 ± 0.1697

      This still gives the same result as when the entire calculation is done at once. Now on the exam you will be given some leeway, and I think your answer would still be within what would be considered the range of correct answers, but please try your calculations again to see if you can find where things went wrong.

  3. Hi jason, i think u just calculate the wrong number in question G)
    My calcularion of lcl is 6.5-(1.96*0.2)=6.108 and ucl is 6.892
    just check am i got wrong?

    1. Keep watching the video for part g. I realize this mistake in my calculations and make the appropriate correction.

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