n = 100
x = 15
σ = 4
Null hypothesis H0: μ = 12.5
Alternative hypotheses Ha: μ ≠ 15.5 (two-tailed test)
The level of significance can be considered as 0.05.
The test statistic
"Z = \\frac{x - \u03bc}{\\frac{\u03c3}{\\sqrt{n}}} \\\\\n\n= \\frac{15 \u2013 12.5}{\\frac{4}{\\sqrt{100}}} \\\\\n\n= \\frac{2.5 \\times 10}{4} \\\\\n\n= 6.25 \\\\\n\nZ_{tab} = 2.33 \\\\\n\nZ> Z_{tab}"
So, we reject H0 at 2% level of significance.
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