Answer to Question #291769 in Statistics and Probability for jgb

Question

In a study of the effects of early Alzheimer’s disease on nondeclarative memory, the Category Fluency Test to establish baseline persistence and semantic memory and language abilities. The eight subjects in the sample had Category Fluency Test scores of 11, 10, 6, 3, 11, 10, 9,11. Assume that the eight subjects constitute a simple random sample from a normally distributed population of similar subjects with early Alzheimer’s disease. (a) What is the point estimate of the population mean? (0.5) (b) What is the standard deviation of the sample? (1) (c) What is the estimated standard error of the sample mean? (0.5) (d) Construct a 90 percent confidence interval for the population mean category

Answer 1

$n=8$

$a)$

The point estimate of the population mean $\mu$ is the sample mean given as, $\bar x={\sum x\over n}={71\over8}=8.875$

$b)$

We first determine the variance given as,

$var(x)=s^2={\sum x^2-{(\sum x)^2\over n}\over n-1}={689-630.125\over7}={58.875\over7}=8.4107$

The standard deviation is,

$sd(x)=\sqrt{var(x)}=\sqrt{8.4107}=2.90$

$c)$

The standard error for the sample mean is given as,

$SE(\bar x)={s\over\sqrt{n}}={2.90\over\sqrt{8}}=1.02534837$

$d)$

A 90% confidence interval for the population mean is given as,

$CI=\bar x\pm t_{{\alpha\over 2},n-1}SE(\bar x)$ where, $t_{{\alpha\over2},n-1}$ is the $t$ distribution table value at $\alpha=0.1$ with $n-1=8-1=7$ degrees of freedom. Thus, $t_{{\alpha\over2},n-1}=t_{{0.1\over2},7}=t_{0.05,7}= 1.894579$ and $CI=8.875\pm(1.894579\times1.02534837)=8.875\pm1.94260349$

Therefore, the 90% confidence interval for the population mean is $(6.93,10.82)$