Null hypothesis "H_0:\\sigma=16."
Alternative hypothesis "H_a:\\sigma<16."
Test statistic: "\\chi^2=\\frac{(n-1)s^2}{\\sigma^2}=\\frac{(15-1)10^2}{16^2}=5.47."
Degrees of freedom: "df=n-1=15-1=14."
P-value: "p=0.022."
(a) Since the P-value is less than 0.05, reject the null hypothesis.
(b) Since the P-value is greater than 0.01, fail to reject the null hypothesis.
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