Unit S3: Statistics 3

A linear combination of independent Normal random variables is also Normal. If$$X\sim N(\mu_x,\sigma_x^2),\qquad Y\sim N(\mu_y,\sigma_y^2)$$independently, then its mean follows the signs in the combination, while each variance contribution uses the square of its coefficient.

Subtraction changes the centre because it reverses Y's contribution to the value. It does not subtract uncertainty: deviations in either variable create spread in the combination, so independent variance contributions add after scaling by squared coefficients.

For independent copies, add one mean and one variance contribution for each copy. If male load $M\sim N(80,100)$ and female load $W\sim N(69,25)$, then the load of six men and three women is$$T=M_1+\cdots+M_6+W_1+\cdots+W_3\sim N(687,675).$$Therefore$$P(T>700)=P\left(Z>\frac{700-687}{\sqrt{675}}\right)=P(Z>0.500\ldots)\approx0.3085.$$

Convert comparisons into one variable before standardising. If $A\sim N(20,9)$ and $B\sim N(8,4)$ independently, then$$D=A-2B\sim N(4,25),$$so$$P(A>2B)=P(D>0)=P\left(Z>\frac{0-4}{5}\right)=P(Z>-0.8)=0.7881.$$

In a simple random sample of size n, every possible set of n distinct population units has the same chance of selection. A complete sampling frame is therefore needed so that every unit can be identified before random selection.

Number the N units consistently, for example 00–69 for 70 students. Read random digits in groups with enough places to represent the largest label. Accept a number only when it is in range and has not already been selected; ignore out-of-range values and repeats until n distinct units have been obtained.

For 70 students labelled 00–69, two-digit random numbers are required. If the stream begins 47, 83, 05, 47, 62, then select 47, ignore 83 because it is outside 00–69, select 05, ignore the repeated 47, and select 62.

The labels identify units but must not influence selection; the random-number source supplies the chance mechanism. Rejecting invalid and repeated labels preserves a sample without replacement while treating eligible labels symmetrically.

Choose a sampling method from the available frame, the population structure and the practical constraints. Stratified, systematic and quota sampling can all spread a sample across a population, but only the first two use a specified probability-based selection rule when carried out correctly.

For population size $N$ and stratum size $N_h$, allocate$$n_h\approx n\frac{N_h}{N},$$then adjust rounding so all allocations total $n$. With 800 employees split 430, 250 and 120 across three cities and sample size 100, proportional allocations are approximately 54, 31 and 15.

For an ordered list of $N=280$ students and sample size $n=40$, the interval is $k=N/n=7$. Choose a random start from 1 to 7, then select that unit and every seventh unit afterwards until 40 are chosen.

An estimator is a rule calculated from a random sample to estimate a population parameter. Its observed value is an estimate. An estimator $\hat\theta$ is unbiased when $E(\hat\theta)=\theta$; its bias is $E(\hat\theta)-\theta$.

The standard error is the standard deviation of an estimator's sampling distribution. It measures how much estimates would vary across repeated samples, rather than how much individual observations vary.

The sample mean $\bar x$ is an unbiased estimate of $\mu$. Dividing the squared deviations by $n-1$, not $n$, makes $s^2$ an unbiased estimate of $\sigma^2$. When $\sigma$ is unknown, $s/\sqrt n$ estimates the standard error of the sample mean.

For the sample 2, 4, 6, $\bar x=4$ and $\sum(x_i-\bar x)^2=8$, so $s^2=8/(3-1)=4$. The estimated standard error of $\bar x$ is $2/\sqrt3$.

For independent observations from a population with mean $\mu$ and variance $\sigma^2$, the sample mean $\bar X$ has the same mean $\mu$ but the smaller variance $\sigma^2/n$. Averaging therefore preserves the centre while reducing sampling variability.

If each observation is Normal, $X\sim N(\mu,\sigma^2)$, then the result is exact: $\bar X\sim N(\mu,\sigma^2/n)$. No proof is required.

If bag weights are $N(50,6^2)$ and $n=36$, then $\bar X\sim N(50,1^2)$. Thus $P(\bar X<48.5)=P(Z<-1.5)\approx0.0668$.

A confidence interval is a range produced by a sampling procedure for an unknown fixed parameter. A 95% procedure is designed so that, over many independent samples, about 95% of the intervals constructed in the same way contain the true parameter.

After one interval has been calculated, the parameter and its endpoints are fixed: the interval either contains the parameter or it does not. The confidence level describes the long-run success rate of the method, not a 95% probability assigned to this particular fixed interval.

For a two-sided test of $H_0:\mu=\mu_0$ at significance level $\alpha$, use the matching $100(1-\alpha)\%$ confidence interval. Reject $H_0$ exactly when $\mu_0$ lies outside the interval; if it lies inside, do not reject $H_0$.

If a 95% confidence interval for $\mu$ is $(22.24,22.56)$, the value 22.50 is compatible with the data at the 5% two-sided level, whereas 23.00 would be rejected.

For a random sample from a Normal population whose variance $\sigma^2$ is known, a two-sided $100(1-\alpha)\%$ confidence interval for $\mu$ is centred on the sample mean and extends by a critical value times its standard error.

Find $\bar x$, calculate $\sigma/\sqrt n$, select the standard Normal critical value for the required confidence level, then subtract and add the margin of error. State both limits with appropriate accuracy.

For $\bar x=22.4$, known $\sigma=0.4$ and $n=36$, a 98% interval uses $z_{0.99}=2.3263$. The margin is $2.3263(0.4/6)=0.1551$, giving $(22.245,22.555)$, or approximately $(22.25,22.56)$.

To test a Normal population mean when $\sigma^2$ is known, compare the observed sample mean with the null value $\mu_0$ in units of its standard error.

State $H_0$ and a one- or two-sided $H_1$ in terms of $\mu$. Calculate $z$, compare it with the appropriate standard Normal critical value (or use its $p$-value), then give a conclusion about $H_0$ and a contextual conclusion.

Suppose $H_0:\mu=1010$ against $H_1:\mu\ne1010$, with known $\sigma=8$, $n=100$ and $\bar x=1008.47$. Then $z=(1008.47-1010)/(8/10)=-1.9125$. Since $|z|<1.96$, do not reject $H_0$ at 5%.

The Central Limit theorem makes the sampling distribution of $\bar X$ approximately Normal for a sufficiently large independent random sample, even when the population distribution is not Normal. This extends Normal confidence intervals and hypothesis tests beyond Normal populations.

When $\sigma^2$ is unknown, use the unbiased sample variance $s^2$ and estimated standard error $s/\sqrt n$. Then use standard Normal critical values for the required large-sample confidence interval or test.

For a large sample with $n=100$, $\bar x=12.4$ and $s=3.0$, an approximate 95% interval is $12.4\pm1.96(3/10)$, giving $(11.812,12.988)$. The same standard error is used to standardise a hypothesised mean.

For two independent Normal populations with known variances, the difference of sample means is Normal. Test a claimed population-mean difference $\delta_0$ by comparing $\bar X-\bar Y$ with $\delta_0$ using the combined standard error.

Independence makes the variances of the two sample means add. The null difference is often 0, but use the value actually claimed and keep the subtraction order consistent in the hypotheses, numerator and conclusion.

With $\bar x=52$, $\bar y=48$, $\sigma_x=6$, $n_x=36$, $\sigma_y=8$, $n_y=64$ and $\delta_0=0$, the standard error is $\sqrt{1+1}=\sqrt2$. Hence $z=4/\sqrt2=2.83$, so reject equality at the 5% two-sided level.

For two independent large samples, the Central Limit theorem allows inference for $\mu_x-\mu_y$ even when the populations are not Normal. If the population variances are unknown, estimate each one from its own sample.

State hypotheses in terms of $\mu_x-\mu_y$, calculate the two-sample standard error from $s_x^2$ and $s_y^2$, standardise, then compare with the correct one- or two-tailed Normal critical value. Interpret the decision in context.

For $n_x=64$, $\bar x=75.2$, $s_x=8$ and $n_y=81$, $\bar y=71.8$, $s_y=9$, testing $H_0:\mu_x-\mu_y=0$ gives standard error $\sqrt{8^2/64+9^2/81}=\sqrt2$ and $z=3.4/\sqrt2=2.40$. This exceeds 1.6449 for a 5% upper-tail test.

A chi-squared test compares observed frequencies $O_i$ with frequencies $E_i$ expected under a null model. Large discrepancies produce a large statistic and evidence against $H_0$; the alternative says the stated model or independence claim does not hold.

For goodness of fit, write $H_0$: the named distribution is a suitable model, and $H_1$: it is not suitable. Required models include discrete uniform, binomial, Normal, Poisson and continuous uniform (rectangular). For a contingency table, write $H_0$: the two named variables are independent (no association), and $H_1$: they are associated.

For goodness of fit, calculate $E_i=Np_i$ from the null distribution. For an $r\times c$ contingency table, $E_{ij}=(\text{row total})(\text{column total})/N$. Compare the statistic with the upper-tail critical value for the chosen significance level and degrees of freedom.

For 50 observations in five equally likely categories, suppose $O=(8,12,9,11,10)$ and $E=(10,10,10,10,10)$. Then $\chi^2=(4+4+1+1+0)/10=1.0$. With 4 degrees of freedom, $1.0<9.488$ at 5%, so do not reject the discrete-uniform model.

Degrees of freedom select the correct chi-squared reference distribution. Count categories after any required combining, then subtract constraints created by the total frequency and by parameters estimated from the same data.

Here $k$ is the final number of cells and $p$ is the number of distribution parameters estimated from the sample. If every parameter is specified in advance, $p=0$. For example, six final Poisson cells with the mean estimated from the data give $\nu=6-1-1=4$. If combining reduces the table to five cells, use $\nu=3$.

The chi-squared approximation requires adequate expected frequencies. When $E_i<5$, combine suitable neighbouring or tail categories before calculating the final statistic; recompute each combined $O$ and $E$, and base $k$ and the degrees of freedom on the combined table.

A $3\times4$ contingency table has $\nu=(3-1)(4-1)=6$. Its expected frequencies still come from row total $\times$ column total $/N$; the observed counts do not determine $\nu$.

Spearman's rank correlation coefficient $r_s$ measures the strength and direction of a monotonic association between two variables by comparing their ranks. It is useful for ordinal data or when the relationship is monotonic but a linear model for the original measurements is unsuitable.

Rank both variables in a consistent direction, calculate $d_i$ as one rank minus the other for each pair, square the differences, and sum them. Reversing both ranking directions changes no differences; reversing only one changes the sign of the coefficient.

For five pairs with ranks $(1,2,3,4,5)$ and $(1,3,2,5,4)$, $\sum d^2=4$. Hence $r_s=1-6(4)/[5(25-1)]=0.8$, showing a fairly strong positive monotonic association.

When values are tied, assign average ranks: each tied value receives the average of the rank positions it occupies. Numerical questions involving ties will not be set, but the correct approach is to calculate the product moment correlation coefficient of the two average-rank lists rather than use the no-ties shortcut unchanged.

A correlation test asks whether a sample coefficient is extreme enough to provide evidence that the corresponding population correlation is not zero. Select the table by coefficient, sample size $n$, significance level and test direction.

For Spearman use $H_0:\rho_s=0$; for product moment correlation use $H_0:\rho=0$. Choose a $>0$ or $<0$ alternative for a predicted direction, or $\ne0$ when either direction counts. A two-sided claim requires the two-tailed table value.

Calculate $r_s$ or $r$ and compare it with the matching critical value. For an upper-tail test reject $H_0$ above the positive critical value; for a two-tailed test reject when the coefficient's absolute value exceeds the critical magnitude.

Suppose $n=7$, $r_s=0.679$ and the 5% upper-tail Spearman critical value is 0.7143. Since $0.679<0.7143$, do not reject $H_0$; there is insufficient evidence of a positive population rank correlation at the 5% level.

Product moment correlation targets linear association in the original values; Spearman targets monotonic association through ranks. Do not interchange their tables.