Answer to Question #109972 in Statistics and Probability for Diana

Question

3. Suppose you are interested in estimating the average salary for biostatisticians in Connecticut. You obtain a random sample of 25biostatisticians hired during the last 5 years. The sample data appeared to be approximately normally distributed with no outliers. The average salary for the sample is $45000 and the sample standard deviation is $3150.

A) Construct a 95%confidence interval for µ the true average salary for newly hired biostatisticians in Connecticut.

B) How large of a sample is needed to estimate µ within $2000?

Answer 1

The provided sample mean is $\bar{X}=45000$ and the sample standard deviation is $s=3150.$

A) The size of the sample is $n=25$ and the required confidence level is 95%.

The number of degrees of freedom are $df=25-1=24,$ and the significance level is  $\alpha=0.05.$

Based on the provided information, the critical t-value for $\alpha=0.05$ and $df=24$ degrees of freedom is $t_c=2.064.$

The 95% confidence for the population mean $\mu$ is computed using the following expression

$$CI=(\bar{X}-{t_c\cdot s \over \sqrt{n}}, \bar{X}+{t_c\cdot s \over \sqrt{n}})$$

Therefore, based on the information provided, the 95 % confidence for the population mean $\mu$ is

$$CI=(45000-{2.064\cdot 3150 \over \sqrt{25}}, 45000+{2.064\cdot 3150 \over \sqrt{25}})=$$

$$=(43699.68, 46300.32)$$

B) The required confidence level is 95%.

$${t_c\cdot s \over \sqrt{n}}\leq2000$$

The critical t-value for $\alpha=0.05$ and $df=11$ degrees of freedom is $t_c=2.200985$

$${t_c\cdot s \over \sqrt{n}}={2.200985\cdot 3150 \over \sqrt{12}}\approx2001.41>2000$$

The critical t-value for $\alpha=0.05$ and $df=12$ degrees of freedom is $tc=2.178813.$

$${t_c\cdot s \over \sqrt{n}}={2.178813\cdot 3150 \over \sqrt{13}}\approx1903.53<2000$$

$n=13$