Test the given claim Assume that a simple random sample is s
Test the given claim. Assume that a simple random sample is selected from a normally distributed population. Use the traditional method in testing hypothesis.
Weights of Women: Supermodels Weights Use a 0.01 significance level to test the claim that weights of female supermodel vary less than the weights of women in general. The standard deviation of weights of the population of women is 29 lb. Listed below are the weights (in pounds) of nine randomly selected supermodels.
125 (Taylor) 119 (Auremann) 128 (Schiffer) 128 (MacPherson) 119 (Turlington) 127 (Hall) 105 (Moss) 123 (Mazza) 115 (Hume)
Heights of Women: Anthropological survey data are used to publish values that can be used in designing products that are suitable for use by adults. According to Gordon, Churchill, et al., women have heights with a mean of 64.1 in. and a standard deviation of 2.52 in.
Use a 0.05 significance level. When designing car seats for women what would be the consequence of believing that the heights of woman vary less than they really vary?
70.8, 66.2, 71.7, 68.7, 67.6, 69.2, 66.5, 67.2, 68.3, 65.6, 63.0, 68.3, 73.1, 67.6, 68.0, 71.0, 61.3, 76.2, 66.3, 69.7, 65.4, 70.0, 62.9, 68.5, 68.3, 69.4, 69.2, 68.0, 71.9, 66.1, 72.4, 73.0, 68.0, 68.7, 70.3, 63.7, 71.1, 65.6, 65.6, 68.3, 66.3
Solution
WEIGHTS OF WOMEN:
Getting the standard deviation of the sample,
s = 7.533259587
Formulating the null and alternative hypotheses,
Ho: sigma >= 29
Ha: sigma < 29
As we can see, this is a left tailed test.
Thus, getting the critical chi^2, as alpha = 0.01 ,
alpha = 0.01
df = N - 1 = 8
chi^2 (crit) = 1.646497373
Getting the test statistic, as
s = sample standard deviation = 7.533259587
sigmao = hypothesized standard deviation = 29
n = sample size = 9
Thus, chi^2 = (N - 1)(s/sigmao)^2 = 0.539833532
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As chi^2 < chi^2(crit), then we REJECT THE NULL HYPOTHESIS.
Thus, there is significant evidence at 0.01 level that weights of female supermodel vary less than the weights of women in general.
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