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A 95% confidence interval for a population mean was reported to be 83 to 87. If $σ=8$, then size of sample is : $(Z_{\alpha /2}=$ approx 2) |
140 120 64 80 |
64 |
The correct answer is Option (3) → 64 Margin of error (ME) = $87 - 85 = 2$ $ME=Z_{\alpha /2}×\frac{σ}{\sqrt{n}}$ $2=1.96×\frac{8}{\sqrt{n}}$ $\sqrt{n}≃8$ $n≃64$ |
