Statistics exercise--please help.?

Question

Two partners were not sure if their samples came from the same Tums tablet. Each student analyzed a sample in triplicate and they obtained the following results. Can they be 98% confident that they analyzed the same tablet?

student I: 14.85, 14.95, 14.71 wgt-% Ca

student II: 15.61, 15.74, 15.80 wgt-% Ca

I don't necessarily need anyone to work this out for me. I just need to know what formula to use? And any explanation for that formula would be nice. This is for a chemistry class but I've never taken statistics before, so I'm confused.

Thanks in advance.

Answer 1

ANSWER: TUMS TABLETS between Student I and Student II are significantly different.

Why??

TESTS BETWEEN TWO MEANS, UNPOOLED T-TEST

1. Parameters of interest:

μ1 = true average of STUDENT I wgt-% Ca TUMS TABLET

μ2 = true average of STUDENT II wgt-% Ca TUMS TABLET

2. Null hypothesis: H0 μ1 - μ2 = 0

3. Alternative hypothesis: Ha μ1 - μ2 ≠ 0 ("two-tailed" test requirement for "significantly different")

4. Test statistic formulae: (Two-sample unpooled t-test)

"t" value and "df" value

http://en.wikipedia.org/wiki/Statistical...

5. Computation of test statistic and df values

SUMMARIZED DATA

Sample  N     Mean   StDev  SE Mean

1       3   14.837   0.121    0.070

2       3  15.7167  0.0971    0.056

Two-Sample T-Test and CI

Difference = mu (1) - mu (2)

Estimate for difference:  -0.880000

98% CI for difference:  (-1.285850, -0.474150)

T-Test of difference = 0 (vs not =): T-Value = -9.85  P-Value = 0.002  DF = 3

6. Determination of the P-value: The test is based on 3 df Table "Look-up" shows area under 3 df curve "two-tailed"

is 0.002.  Because P-value < α (=1 - 0.95)/2 [0.002 < 0.025] Null Hypothesis must be rejected. Null Hypothesis of no difference between TUMS TABLET of Student I and Student II must be rejected at a confidence level of 98%.  TUMS TABLETS between Student I and Student II are significantly different.