11: Variances and means (Key even)

11.2 Leaves on a common stem. Based on these plots, compare group means and variances. (Calculations not necessary.)

(A)

0|9| |8| 0|7|0 |6|0 0|5|0 |4|0 0|3|0 |2| 0|1| �10

(B)

0|9| 0|8| 0|7|0 0|6|0 0|5|0 |4|0 |3|0 |2| |1| �10

(C)

0|9|0 |8|0 0|7|0 |6|0 0|5|0 |4| 0|3| |2| 0|1| �10

Means equal

Group 1 mean greater

Group 1 mean smaller

Group 1 variance greater

Variances equal

Grou p1 mean greater

11.4 Particulate matter in air samples. units = micrograms/m³, n₁ = 8, n₂ = 8

(A) Create stemplots on a common stem to compare these two distributions. Use split stem-values for your plot. Below. Discuss your findings. Readings from the sites have similar central locations (median locations underlined), but Site 1 has much greater variance. Site 1 has a high outlier.

Site 1| |Site 2
---------------
    42|2|
     8|2|
     2|3|234
    86|3|6689
     2|4|0
      |4|
      |5|
      |5|
      |6|
     8|6|
     (�10)

(B) Calculate the means and standard deviations of the data from the two sites.
Site 1: xbar = 36.25, s = 14.56         Site 2: xbar = 36.00, s = 2.88.
How do these summary statistics complement your stemplot analysis? confirm sites have equal locations but different variability.

(C) Test the variances for inequality with an F ratio test. Include a statement of the null hypothesis.
H₀: sigma₁² = sigma₂²         F_stat = 14.56² / 2.88² = 25.56 with df₁ = 7 and df₂ = 7        P < 0.001 (P = 0.00018), showing variances differ significantly.

(D) Would you use a pooled (equal variance) t procedure to test of means? No
Explain your response. The differ too much; pooled (equal variance) t test assumes one underlying variance when calculating s²pooled

(E) Test the means for inequality. Show all hypothesis testing steps.
H₀: �₁ = �₂         t_stat = 0.05 with df_conserv = 7        P > 0.40, showing difference is not significant        Here's the SPSS output:

(F) What is the most important finding in this analysis? ...high variability and high outlier at Site 1. That high pollution days can have serious health consequences for cardio-resp compromised individuals. Was the test of means revealing or obscuring? Obscuring.

11.12 Calcium supplementation on blood pressure. For the data in Exercise 11.11, test the decreases in blood pressure for a significant difference. Summary statistics must first be calculated.
Treatment group (n₁ = 10) has a mean decline of 5.00 mm Hg (s₁ = 8.743 Hg) Placebo group (n₂ = 11) mean decline of −0.27 (s₂ = 5.901).
In testing H₀: �₁ = �₂ vs. H₁: �₁ ≠ �₂; t_stat = 1.634; df_Welch = 15.591; P = 0.129 (two-sided). Working by hand, df_conser = 9, P = 0.14. Either way, no significant difference in changes in blood pressure.
Note small sample size. data may be too sparse to detect a statistically significant difference. Maybe more study is needed.