Question #328066

Find 80%, 85%, 99, 99% confidence interval for the sample mean of a population,. if we know that a random sample of 70. people from the population, the sample mean is 60 and standard deviation. is. 5

1
Expert's answer
2022-04-15T03:09:06-0400

n=70xˉ=60s=5Confidenceinterval:(xˉsnt1+γ2,n1,xˉ+snt1+γ2,n1)=(600.59761t1+γ2,69,60+0.59761t1+γ2,69)γ=0.8:(600.597611.2939,60+0.597611.2939)=(59.227,60.773)γ=0.85:(600.597611.4557,60+0.597611.4557)=(59.130,60.870)γ=0.9999:(600.597614.1304,60+0.597614.1304)=(57.532,62.468)n=70\\\bar{x}=60\\s=5\\Confidence\,\,interval:\\\left( \bar{x}-\frac{s}{\sqrt{n}}t_{\frac{1+\gamma}{2},n-1},\bar{x}+\frac{s}{\sqrt{n}}t_{\frac{1+\gamma}{2},n-1} \right) =\left( 60-0.59761t_{\frac{1+\gamma}{2},69},60+0.59761t_{\frac{1+\gamma}{2},69} \right) \\\gamma =0.8:\\\left( 60-0.59761\cdot 1.2939,60+0.59761\cdot 1.2939 \right) =\left( 59.227,60.773 \right) \\\gamma =0.85:\\\left( 60-0.59761\cdot 1.4557,60+0.59761\cdot 1.4557 \right) =\left( 59.130,60.870 \right) \\\gamma =0.9999:\\\left( 60-0.59761\cdot 4.1304,60+0.59761\cdot 4.1304 \right) =\left( 57.532,62.468 \right)


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