Population standard deviation is unknown , we will use t test
Degrees of freedom "=n-1=25-1=24"
Critical value=
"t_{\u03b1\/2, df} = t_{0.05\/2.24} = t_{0.025, 24} = \u00b12.0639"
95% confidence interval is given by
"\\bar{x} \u00b1 t_{\u03b1\/2, df} \\times \\frac{s}{\\sqrt{n}} \\\\\n\n= (10.98 -2.0639 \\times \\frac{0.604}{\\sqrt{25}} < \\mu < 10.98 + 2.0639 \\times \\frac{0.604}{\\sqrt{25}}) \\\\\n\n= (10.98 -0.24932 < \\mu < 10.98 + 0.24932) \\\\\n\n= (10.73068 < \\mu < 11.22932)"
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