Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Confidence Intervals

  • 26-05-2009 3:55pm
    #1
    mcdonagh521
    Closed Accounts Posts: 9 ✭


    Hello.
    I'm having trouble with this question. I can do part (a) but i don't no where to start on question (b) and (c). Can anyone help me?

    In a quality assurance study, 10 bottles of lemonade, each nominally containing 1 litre, are selected and the true volume of lemonade contained in each is measured.
    The results, in mls, are
    997; 1003; 999; 988; 1006;
    1005; 1007; 1009; 994; 1002:

    (a) Calculate the sample mean and sample standard deviation for these measurements.

    (b) Construct the 99% confidence interval for the true mean using a distribution appropriate for this small sample.

    (c) A further study is undertaken, in which 150 sample bottles are used. The sample mean and standard deviation are found to be x = 998 and s = 4 respectively.
    Using an appropriate distribution, calculate the 99% confidence interval for the true mean.

    Any help would be appreciated, thanks.


Comments

  • #2
    gra26
    Registered Users, Registered Users 2 Posts: 197 ✭✭gra26


    Ok, Well in part b since your n is so small(< 30) you need to use the student t-tables to get certain values. in part c you use the normal tables to get similar values.

    SE: Standard error = (standard deviation)/(square root of n)

    A confidence interval is given by: mean + and - (????)*SE

    The ???? is going to change depending on what % Confidence interval you're looking for and whether you're using the t or normal tables.

    To find ????:
    For part c you need to find the positive z value such that between + and - z there lies 99% of the area under the curve. Then just use the formula. Obviously this changes as the % you use changes.

    As for part b, If you're using a 2-tailed t-table to get the value what you need to use are (a) the degrees of freedom (n-1) and (b)100-99=1 for the column and so find the value where column and row intersect.
    If you're using a 1-tailed t-table you need (a) degrees of freedom again and also (b) an alpha/2 value. alpha in this case is 1-0.99=0.01 so you need to use 0.01/2 and again find the value where row and column intersect.


  • #3
    mcdonagh521
    Closed Accounts Posts: 9 ✭mcdonagh521


    Thanks for your reply.
    I understand most of it just not how to work out the (????).
    Can you show me how to work it out with this example?


  • #4
    gra26
    Registered Users, Registered Users 2 Posts: 197 ✭✭gra26


    that any better?


  • #5
    MathsManiac
    Registered Users, Registered Users 2 Posts: 1,595 ✭✭✭MathsManiac


    I think gra26 had a bit of a typo there: for part (b) the degrees of freedom is 10-1=9.

    So, the (????) is the two-tailed 1% critical value for the t-distribution (i.e. the 0.005 ordinate for the t-distribution) with 9 degrees of freedom. Look it up and tell us what you got.

    In part (c), you need to find the value of z that has an area of .995 to the left of it.


Advertisement