"1) N_1=36, \\overline{x}_1=5.2"
"H_0: a=a_0=5, H_1:a>a_0=5\\text{ (one-sided)}\\\\\na \\text{ is population mean}\\\\\n\\alpha=0.01\\\\\n\\sigma=0.15\\\\\n\\text{We assume that volume has normal distribution with mean } a\\\\\n\\text{and standard deviation } \\sigma=0.15.\\\\\n\\text{We will use the following random variable:}\\\\\nU=\\frac{(\\overline{X}-a_0)\\sqrt{n}}{\\sigma}\\\\\n\\overline{X}\\text{ --- random value of sample mean}\\\\\nu_{obs}=\\frac{(5.2-5)\\sqrt{36}}{0.15}=8\\\\\n\\Phi(u_{cr})=\\frac{1-2\\alpha}{2}=0.49\\\\\nu_{cr}=2.33\\\\\n\\Phi(x)\\text{ --- Laplace function}\\\\\n(2.33,\\infty)\\text{ --- critical region}\\\\\nu_{obs} \\text{ is in the critical region. So we reject } H_0.\\\\\n\\text{We cannot be 99 percent sure that the sample have come from}\\\\\n\\text{a population of 5 litres}." "2) N_2=36, \\overline{x}_2=4.95\\\\\nH_0: a=a_0=5, H_1:a<a_0=5\\text{ (one-sided)}\\\\\na \\text{ is population mean}\\\\\n\\alpha=0.01\\\\\n\\sigma=0.15\\\\\n\\text{We assume that volume has normal distribution with mean } a\\\\\n\\text{and standard deviation } \\sigma=0.15.\\\\\n\\text{We will use the following random variable:}\\\\\nU=\\frac{(\\overline{X}-a_0)\\sqrt{n}}{\\sigma}\\\\\n\\overline{X}\\text{ --- random value of sample mean}\\\\\nu_{obs}=\\frac{(4.95-5)\\sqrt{36}}{0.15}=-2\\\\\n(-\\infty,-u_{cr})\\text{ --- critical region where } u_{cr}=2.33\\\\\nu_{obs} \\text{ is not in the critical region. So we accept } H_0.\\\\\n\\text{We can be 99 percent sure that the sample have come from}\\\\\n\\text{a population of 5 litres}."
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