A set of 312 small wild birds caught at random was found to be composed of 102 males (M)?

A set of 312 small wild birds caught at random was found to be composed of 102 males (M), having sample mean wing length = 12.5 cm, with a standard deviation of 0.81 cm, and 210 females (F), with sample mean wing length 11.75 cm, and a standard deviation of, 0.74 cm. We propose to use this data to find out whether there is a difference between the male and female birds wing lengths








(a) Establish the null, and alternative hypotheses. Find the estimated standard error, ESE, for the sampling distribution.











(b) Find the two sample test statistics, Z, associated with M, and, F.











(c) Do you accept or reject the null hypothesis at the 10%, significance level?

A set of 312 small wild birds caught at random was found to be composed of 102 males (M)?
LARGE SAMPLE, TWO MEANS TEST, INDEPENDENT SAMPLES, NORMAL


DISTRIBUTION - SIX STEP PROCEDURE





1. Parameters of interest: "μ1" = "small wild birds, males, wing length " population mean, "μ2" = "wild birds, females, wing length" population mean





2. Null Hypothesis H0: (μ1 - μ2) = 0 (no difference between means)





3. Alternative Hypothesis H1: (μ1 - μ2) %26lt;%26gt; 0 (different means)





4. Sample Data "males" x1-bar = sample mean [12.5] s1=sample standard deviation [0.81]


n1=number of samples [102] "females" x2-bar = sample mean [11.75] s2=sample standard


deviation [0.74] n2=number of samples [210]





5. TEST STATISTICS from "look-up" Table Standard Normal Distribution


z1(males) = (12.5-population mean)/ 0.81 = ???


z2 (females) = (11.75-population mean)/ 0.74 = ???





H0: (μ1 - μ2) = 0 [CANNOT BE DETERMINED WITHOUT KNOWLEDGE OF "population mean"]





H1: (μ1 - μ2) ] %26lt;%26gt; 0 [CANNOT BE DETERMINED WITHOUT KNOWLEDGE OF "population mean"]





6: Conclusion: Both Null Hypothesis H0: (μ1 - μ2) = 0 and Alternate Hypothesis H1: (μ1 - μ2) ] %26lt;%26gt; 0 CANNOT BE DETERMINED WITHOUT KNOWLEDGE OF "population mean"].