Course - Morgan Park High School



Name:__________________________ Introduction to T-Tests

The difference between a Z test and a T test:

• To perform a Z test we must know the standard deviation of the population, sigma ( σ ).

• For a T test we use the standard deviation of the sample ( s ).

• The standard deviation of a sample varies from sample to sample. This is not the case for sigma.

• We must use the T table for a T-test because the sample standard deviation varies.

• For a T-test, we must also determine the degrees of freedom

• For a single-sample test, the degrees of freedom equals the sample size minus one [pic].

Example:

When working properly, A soda machine will dispense soda with a mean [pic] = 12 ounces (note: we are not told sigma, so we will perform a T-test). A quality control engineer is concerned that the machine is under filling the cups. He draws a simple random sample of 8 cups and records the following amounts:

|12.12 |11.35 |11.51 |11.86 |11.75 |11.45 |11.92 |12.22 |

1) The engineer conducts a test of significance of :

[pic]

[pic]

2) Put the data in a list. Run 1-Var Stats.

a) What does x-bar equal? __________________________________

b) What does the sample standard deviation equal? (Note: Look for Sx not σ) ____________________

3) A T-test is similar to a T-test except we use S instead of σ

[pic] [pic]

a) Plug the appropriate values into the T-test formula and find the test statistic (T-statistic).

b) Calculate your degrees of freedom. [pic]

c) Find a copy of the T-table (Table C at the back of your book or from a handout.)

d) In the table, the left-hand column denotes the degrees of freedom. Find the appropriate row.

e) Scan across the row until you find the two values your T-statistics falls between.

i) If you have done this correctly so far, your T-statistic should be -2.03

ii) When looking in the T-table we are only concerned with the absolute value of our test statistic. So, we’ll look for 2.03 in row 7 (because the degrees of freedom equals 8 – 1 or 7).

iii) 2.03 falls between the values 1.895 and 2.365

iv) Look to the top of the two columns. The p-value that corresponds to our statistic falls between these two values (.05 and .025)

v) So, using the table we can approximate that the p-value is greater than 2.5% but less than 5%

4) To get a more exact p-value, use the calculator. STAT>TESTS>2:T-Test

You should use the calculator for these tests, but you should be familiar with how the table works.

5) What is your decision regarding the null hypothesis? Is the machine significantly under filling cups? Assume the significance level is 5%.

Problem #2

The composition of the earth’s atmosphere may have changed over time. One attempt to discover the nature of the atmosphere long ago studies the gas trapped in bubbles inside ancient amber. Amber is tree resin that has hardened and been trapped in rocks. The gas in bubbles within amber should be a sample of the atmosphere at the time the amber was formed. Measurement on specimens of amber from the late Cretaceous era (75 to 95 million years ago) gives these percents of nitrogen:

63.4 65.0 64.4 63.3 54.8 64.5 60.8 49.1 51.0

These values are quite different from the present 78.1% (0.781) of nitrogen in the atmosphere, but are these differences significant? Assume (this is not yet agreed on by experts) that these observations are an SRS from the Cretaceous atmosphere.

1) State the null and alternative hypotheses

2) Calculate the test statistic and degrees of freedom.

3) Find the p-value in the table & also in the calculator. Draw a sketch.

4) What is your decision regarding the null hypothesis? Has the percentage of nitrogen significantly changed? Assume the significance level is 5%.

The developer of a new filter for filter-tipped cigarettes claims that it leaves less nicotine in the smoke than does the current filter. Because cigarette brands differ in a number of ways, he tests each filter on one cigarette of each of nine brands and records the difference between the nicotine content for the current filter and the new filter. The mean difference is [pic] = 1.32 milligrams (mg), and the standard deviation of the differences is s = 2.35 mg.

1. Describe the population of interest and the parameter for which inference is being performed.

2. State H0 and Ha for this study in both symbols and words.

3. What conditions are required to carry out the significance test? Discuss the validity of each.

4. Determine the test statistic and the P-value. Show your work.

5. What do you conclude?

6. A 90% confidence interval for the mean amount of additional nicotine removed by the new filter is (–0.14, 2.78). Determine t* for this interval. Then interpret the interval.

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