Answer to Question #94931 in Statistics and Probability for Marian

Question

a statistics practitioner randomly sampled 100 observations from a population whose standard deviation is 5 and found that sample mean is 10. Estimate the population mean with 60% confidence.
b. Repeat part a with a sample size of 25
c. Repeat part a with a sample size of 10
d. describe what happens to the confidence interval estimate when the sample size decreases

Answer 1

If we know the standard deviation of this population, the confidence interval for the mean can be found as

$$(\bar X - {q_{1 - \frac{\alpha }{2}}}\frac{\sigma }{{\sqrt n }},\bar X + {q_{1 - \frac{\alpha }{2}}}\frac{\sigma }{{\sqrt n }})$$

where ${\bar X}$ - sample mean, $n$ - sample size, $\sigma$ - known standard deviation and ${q_{1 - \frac{\alpha }{2}}}$ is quantile of the standard normal distribution of level ${1 - \frac{\alpha }{2}}$ . Our significance level is

$$\alpha = 1 - \frac{C}{{100\% }} = 1 - \frac{{60\% }}{{100\% }} = 0.4$$

so we need to find ${q_{1 - \frac{\alpha }{2}}} = {q_{1 - \frac{{0.4}}{2}}} = {q_{0.8}}$ quantile. To do so use it's definition

$$\Phi ({q_y}) = y$$

where $\Phi (x)$ is the CDF function of the standard normal distribution. Using the table of this CDF (I used the table from six-sigma-material.com)

We need to find a value of independent variable z such gives $0.8$ as result. From the table we see, the value of z lies approximatelty at $z = 0.84$ (for it $\Phi (0.84) = 0.7995$). If we don't need high accuracy we don't use interpolation and just approximately accept

$${q_{0.8}} \approx 0.84$$

a) Using values we got above with $n=100$

$$(\bar X - {q_{1 - \frac{\alpha }{2}}}\frac{\sigma }{{\sqrt n }},\bar X + {q_{1 - \frac{\alpha }{2}}}\frac{\sigma }{{\sqrt n }}) \approx (10 - 0.84 \cdot \frac{5}{{\sqrt {100} }},10 + 0.84 \cdot \frac{5}{{\sqrt {100} }}) = (9.58,10.42)$$

b) Do the same with $n=25$

$$(\bar X - {q_{1 - \frac{\alpha }{2}}}\frac{\sigma }{{\sqrt n }},\bar X + {q_{1 - \frac{\alpha }{2}}}\frac{\sigma }{{\sqrt n }}) \approx (10 - 0.84 \cdot \frac{5}{{\sqrt {25} }},10 + 0.84 \cdot \frac{5}{{\sqrt {25} }}) = (9.16,10.84)$$

c) Do the same with $n=10$

$$(\bar X - {q_{1 - \frac{\alpha }{2}}}\frac{\sigma }{{\sqrt n }},\bar X + {q_{1 - \frac{\alpha }{2}}}\frac{\sigma }{{\sqrt n }}) \approx (10 - 0.84 \cdot \frac{5}{{\sqrt {10} }},10 + 0.84 \cdot \frac{5}{{\sqrt {10} }}) \approx (8.67,11.33)$$

d)Confidence interval for the mean is increasing with decreasing size of population.

P.S. Note that if we don't know standard deviation for population we need to use other method.