Question

Samples (X1, X2, …. XN ) are taken from a normal distribution with unknown population mean and population variance. The sample mean is 10.5 and the sample variance is 5.5. What is the 95% confidence interval for the population mean?

Answer 1

Answer: 95% confidence interval would be

$(10.5-\dfrac{10.78}{\sqrt{N}},10.5+\dfrac{10.78}{\sqrt{N}})$

Step-by-step explanation:

Since we have given that

Sample mean = 10.5

Sample variance = 5.5

Sample size = N

We need to find the 95% confidence interval for the mean.

z = 1.96

So, the confidence interval would be

$\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=10.5\pm 1.96\dfrac{5.5}{\sqrt{N}}\\\\=(10.5-\dfrac{10.78}{\sqrt{N}},10.5+\dfrac{10.78}{\sqrt{N}})$

Hence, 95% confidence interval would be

$(10.5-\dfrac{10.78}{\sqrt{N}},10.5+\dfrac{10.78}{\sqrt{N}})$