Null hypothesis "H_0:\\mu=70."
Alternative hypothesis "H_a:\\mu\\ne70."
Test statistic:
"z=\\frac{\\bar x-\\mu}{\\frac{\\sigma}{\\sqrt{n}}}=\\frac{75-70}{\\frac{5}{\\sqrt{200}}}=14.14."P-value: "p=2P(Z<-14.14)<0.0001."
Since the P-value is less than 0.05, reject the null hypothesis.
It is safe to conclude at 0.05 significance level that the sample is significantly different from the population.
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