Since the population standard deviation is known, Z-test is used.
"H_0:\\mu=100"
"H_1:\\mu\\ne100"
Let "\\alpha=0.05"
"Z=\\frac{\\bar{X}-\\mu}{\\frac{\\sigma}{\\sqrt{n}}}"
"Z_{0.05}=1.96" (two tailed)
"Z=\\frac{140-100}{\\frac{15}{\\sqrt{30}}}=14.61"
Since "Z=14.61>1.96" , we reject the null hypothesis and conclude that the raw cornstart had an effect at 5% level of significance.
Comments
Leave a comment