A set of 312 small wild birds caught at random was found to be?

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?
Here we have samples derived from two populations (male and female birds) and wish to compare the means of the two populations to determine whether they differ or not.





The null hypothesis is that μ1 = μ2 (μ1 is the mean for the male population, μ2 is the mean for the female population), and the alternative hypothesis is that μ1 %26gt; μ2.





The expected value for the difference in the sample means, (m1 - m2), is simply (μ1 - μ2), and the standard error of the difference in sample means is √{ (σ1²/ n1) + (σ2²/n2) }





Using the sample standard deviations as estimates for the population standard deviations, we have, therefore,





Standard error of the difference = √{ (0.81² / 102) + (0.74² /210) } = √0.00904 = 0.095





The 90% z-value is 1.28 standard deviations : here we have (m1 - m2) / S.E. = (12.5 - 11.75) / 0.095, which is equal to 7.89, and is greater than 1.28, therefore the null hypothesis is refuted : there is a difference in the wing-lengths.


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Or, are you dealing just with the distributions of the individual populations, and not with the difference between the two ?





In this case, the standard error of the mean for a single sample = σ / √n, so for the male population we have





S. E. = 0.81 / √102 = 0.080, and for the female population S.E. = 0.74 / √210 = 0.051





The null hypothesis is that m1 = m2, the alternative hypothesis is that m1 %26gt; m2.





Considering the male population, the probability of a wing-length of 11.75 is given by Φ(z), where z = (12.5 - 11.75 ) / 0.08 = 9.375.





Considering the female population, the probability of a wing-length of 12.5 is given by Φ(z), where z = (11.75 - 12.5) / 0.051 = -14.7





As stated above, the z-value which corresponds to a significance level of 10% is 1.28 (either + or - ), and these z-values are well outside this, confirming the difference in the two means.
Reply: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"].