A researcher claims that a 20 year old women on a special diet will have an average weight of 110 pounds. A sample of 15 women has an average weight of 112.5 lbs and a standard deviation of 5 pounds. At 8 = 0.01 can the clean be rejected?
The following null and alternative hypotheses need to be tested:
"H_0:\\mu=110"
"H_a:\\mu\\not=110"
This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.
Based on the information provided, the significance level is "\\alpha = 0.01," "df=n-1=14" degrees of freedom, and the critical value for a two-tailed test is "t_c =2.976842."The rejection region for this two-tailed test is "R = \\{t:|t|> 2.976842\\}."
The t-statistic is computed as follows:
Since it is observed that "|t| = 1.9365<2.976842=t_c," it is then concluded that the null hypothesis is not rejected.
Using the P-value approach:
The p-value for two-tailed "df=14" degrees of freedom, "t=1.9365" is "p=0.073257," and since "p=0.073257>0.01=\\alpha," it is concluded that the null hypothesis is not rejected.
Therefore, there is not enough evidence to claim that the population mean "\\mu" is different than 110, at the "\\alpha = 0.01" significance level.
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