Statistics Test Review 2: Probability, Sampling Distributions, and Confidence Intervals

Vocabulary practice flashcards covering standard error formulas, normal probabilities, point estimates, confidence intervals, sample size calculations, and TI-84 procedures.

Inverse Normal Value (Question 8)

The cutoff value found using the TI-84 function invNorm to locate the height separating the tallest $5\%$ of females in a normal distribution with a mean of $65\,\text{inches}$ and standard deviation of $2.5\,\text{inches}$, calculated at area $1 - 0.05 = 0.95$ to give $69.11\,\text{inches}$.

Standard Error of the Mean (Question 12)

The standard deviation of the sampling distribution of sample means, computed as $SE = \frac{\sigma}{\sqrt{n}}$; for a population standard deviation $\sigma = 10$ and sample size $n = 16$, it equals $\frac{10}{\sqrt{16}} = \frac{10}{4} = 2.5$.

Sampling Distribution Probability for Professor Ages (Question 13)

The probability that the average age of $n = 16$ randomly selected professors is less than $48\,\text{years}$, where the population mean is $45.0\,\text{years}$ and standard deviation is $6\,\text{years}$ ($SE = \frac{6}{\sqrt{16}} = 1.5$), found via TI-84 normalcdf to be $0.9772$.

Sampling Distribution of a Die Roll Mean (Question 14)

The probability $P(\bar{x} > 3)$ for a fair six-sided die with mean $3.5$ and standard deviation $1.71$ rolled $n = 30$ times, computed using standard error $SE = \frac{1.71}{\sqrt{30}} = 0.3122$ and TI-84 normalcdf as $0.9453$.

Point Estimate (phat) (Question 15)

The sample proportion used as an estimate for the population proportion $p$, defined by the formula $\hat{p} = \frac{x}{n}$; for $x = 385$ economy-focused voters out of $n = 500$, it is $\frac{385}{500} = 0.77$.

Point Estimate from a Confidence Interval (Question 19)

The center value of a confidence interval, calculated by averaging the lower and upper bounds: $\frac{82 + 88}{2} = \frac{170}{2} = 85$.

Margin of Error from Confidence Limits (Question 20)

Half the width of a confidence interval, calculated for an interval between $72\%$ ($0.72$) and $76\%$ ($0.76$) as $E = \frac{0.76 - 0.72}{2} = \frac{0.04}{2} = 0.02$.

Confidence Interval Construction from Margin of Error (Question 21)

The interval obtained by adding and subtracting the margin of error from the point estimate; for a point estimate of $0.82$ and margin of error $0.03$, the $90\%$ confidence interval is $(0.82 - 0.03, 0.82 + 0.03) = (0.79, 0.85)$.

1-PropZInt (Question 22)

The TI-84 calculator procedure used to construct a confidence interval for a single population proportion given $x = 268$ distracted drivers out of $n = 400$ non-fatal accidents ($\hat{p} = 0.67$) at a confidence level of $0.95$.

Sample Size Determination for Proportions (Question 23)

The minimum sample size formula $n = \frac{z^2 \cdot p(1-p)}{E^2}$; calculated for a $95\%$ confidence level ($z = 1.96$), worst-case proportion $p = 0.50$, and margin of error $E = 0.02$ as $n = \frac{(1.96)^2 \cdot (0.50)(0.50)}{(0.02)^2} = \frac{0.9604}{0.0004} = 2401$.

TInterval (Question 24)

The TI-84 calculator command used to calculate a confidence interval for a population mean given sample statistics $\bar{x} = 90$, sample standard deviation $S_x = 20$, sample size $n = 16$, and confidence level $0.95$, yielding $(79.3, 100.1)$.

Margin of Error for a Proportion Formula (Question 25)

The formula $E = z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, evaluated for a $95\%$ confidence interval ($z = 1.96$) where $x = 300$ out of $n = 500$ residents oppose raising taxes ($\hat{p} = 0.60$) to yield $E = 1.96\sqrt{\frac{0.60(0.40)}{500}} \approx 0.04$.