ACTIVITY SET 1 Jan - Penn State Department of Statistics

By Carol Mitchell,2014-11-08 07:18
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ACTIVITY SET 1 Jan - Penn State Department of Statistics

Confidence Intervals Proportion and One Mean

    1 The term sampling frame refers to the group that actually had a chance to get into the sample. Ideally, this is the same as the population of interest, but sometimes it isn‟t. In the following situation, describe

    the population, the sampling frame, the sample, the parameter of interest, and the statistic.

    A Gallup Poll is done using random digit dialing to reach individuals in households with land-line telephones. The purpose is to estimate the proportion of U.S. adults who favor stronger gun control laws. One-thousand persons are sampled, and 63% favor stronger gun control.

a. Population =

b. Sampling frame =

c. Parameter =

d. Sample =

e. Statistic =

2 From the Data Sets folder open the Class Survey data and select a column of your choice to answer

    the questions following the special notes.

Special Notes:

    1. To perform a proportion interval your column can only have two options (e.g. Male/Female for Gender, Yes/No for smoking, etc.) However, you can manipulate your data to create a column with just two outcomes. For example, if you wanted to manipulate the GPA data to either ? 3.00 or < 3.00 you can

    use Calc > Calculator, enter a new variable name in the text box “Store result in variable” and then

    in the expression Window enter „GPA‟ ? 3.00. This will create in your new variable name column a

    series of „1‟ and „0‟ where the 1‟s are for those with a 3.00 GPA or higher and 0‟s for those less than 3.00.

    2. When Minitab calculates a proportion, if using variables with Text responses (e.g. Male/Female; Yes/No) the proportion will be calculated on the second word listed alphabetically (i.e. Male or Yes). You will see this in the output where it reads “Event = Male” or “Event = Yes”. So if you want Female

    or No you will have to refer to Special Note 1.

a. What column of data did you select?

b. What would be your population of interest?

c. How is the class survey representative of that population?

    d. How would you best describe the sampling technique used in attaining the Class Survey?

    e. Define a parameter value of interest (e.g. from Activity 9 the U.S. Government reported that 23% of US adults ages 18-24 smoked cigarettes). If you cannot easily find one think of one. For instance (although you cannot use this parameter value now since I am giving it to you!), maybe you believe that 25% of college students cheat or have cheated on a significant other.


    ˆf. Referring to condition regarding using the normal approximation for sample proportions (i.e. is n ? 5 p

    ˆand n(1-) ? 5) verify that this condition has been met thus allowing you to select in Minitab the Option p

    Use test and interval based on Normal Distribution

g. Use Minitab Stat > Basic Statistics > 1-Proportion to create a 90% 1-proportion confidence interval

    on the column you selected in part “a” above. Remember to check Option Use test and interval based

    on Normal Distribution . What is your interval?

h. For your interval, answer the following:

    ˆ What is the sample proportion, ? p

     What is the z multiplier used for your interval?

     What is the standard error?

     What is the margin of error?

    i. What is the sample size? The number of students who submitted a survey was 226. If your sample size is not 226 why do you believe your number differs?

j. List the value and expression for the following:



     Sample Size:

    k. If the confidence level in part g were changed to 95% would your resulting interval be wider or narrower?

3 For mean confidence intervals we call into use T-Table which can be found in this week‟s folder. The

    first concept to understand is the idea of Degrees of Freedom (DF). For our activity today, since we are only concerned with one mean (either from one sample of a difference between paired data), DF = sample size 1 (i.e. n 1). When finding confidence intervals, T-Table provides the t multiplier (t*) for the

    confidence interval expression:


    n if data is consists of only one sample

    NOTE: you will notice that the DF column in this tables only increases by 1 from 1 to 30. After that the increments vary. If your DF is NOT found in the table then conservatively use the CLOSEST


    df in the table that does not exceed the DF of interest. For example, if the degrees of freedom were 37 then from the table use 30.

     Find t-multipliers and DF from Table A2 for the following conditions:

Confidence Level 90%, n = 25: t* = DF =

Confidence Level 95%, n = 25: t* = DF =

Confidence Level 95%, n = 35: t* = DF =

Confidence Level 99%, n = 35: t* = DF =

    What do you notice that happens to t* as the level of confidence increases for the same sample size?

    What do you notice that happens to t* as sample size increases for the same level of confidence?

4 Using the Class Survey data, PSU wants to estimate the true SATM scores for its undergraduate

    population. Assuming that our survey represents this population, calculate a 1-mean 95% confidence interval to estimate the parameter. First we will do by hand and then using Minitab. To start, the sample

    xmean () is 599 and n = 216 and s = 85.34.

a. Calculate the standard error of the mean.

b. What are the DF and t* from T-table?

DF = t* =

    c. Calculate the 95% Confidence Interval and provide an interpretation of this interval.



    d. Now use Minitab to verify your results. Go to Stat > Basic Statistics > 1-Sample t. For “Sample in

    columns” enter SATM. Click Options and be sure that 95 is entered as the Confidence Level and “Not

    Equal” is the Alternative (if you don‟t have Not Equal then Minitab will produce an interval with just either a lower bound or upper bound.) Click OK twice and copy and paste your Minitab results. Do your results by hand and those from Minitab roughly match?

e. Write a sentence that explains what this interval means.

    f. Based on the interval calculated, if Penn State claimed that the true mean SATM score for the 2005 incoming freshman was 610 would you believe them? Explain.


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