Sample size determination

Experiments were performed to establish the resistance of various steel alloys to corrosion.

Experiments performed for a certain alloy showed that this alloy does not succumb to corrosion for 875 days on average. This determination is accompanied by a standard deviation equal to 85 days.

Using the probability integral, indicate how many sheets manufactured from this alloy must be subjected to testing so that, with probability 0.90, one can expect the deviation from the mean to be no more than $5\%$.

Steel wire used in the manufacture of cables withstands, on average, a tensile force of $6720 \text{ kg/cm}^{2}$. The standard deviation of individual samples from this mean has the value $220 \text{ kg/cm}^{2}$. A series of successive observations showed that, with respect to the tensile forces withstood by individual wire samples, a normal distribution law holds.

From each batch of wire arriving at the plant, 20 specimens were subjected to testing. The test results are given in the table below.

Value of the tensile force withstood by the wire per 1 cm2

Is it possible, on the basis of these tests, to assert that both batches of wire are of the required quality and that the test results should be explained by chance of sampling?

The distribution of values of the quantity $x$ has standard deviation $\sigma$. An individual measurement of the values of $x$ is accompanied by a mean square error $\sigma '$. To measure $x$ one must use specimens, the manufacture of which costs $a$ rubles apiece. The cost of carrying out one measurement is $a'$ rubles. How many individual specimens should be manufactured, and how many measurements should be made on each, so that the mean value of the quantity $x$, established as a result of this series of observations, is accompanied by a mean square error not exceeding $\alpha$, and the cost of carrying out the experiment is as small as possible?

How many independent tosses of a fair coin are required for the probability that the average number of heads differs from 0.5 by less than $1 \%$ to be at least 0.99 ?

Three candidates $A, B$, and $C$ are running for President. In order to predict the outcome of the election, a number of people, each selected at random and independently of the others, are asked their choice. The predicted percentage $p_{X}$ of votes for a each candidate $X \in \{ A, B, C\}$ is worked out by dividing the number of people in the sample who are going to vote for $X$ by the total number of people in the sample. How many people need to be in the sample to substantiate the following claim: For each candidate $X$, the probability that the predicted percentage $p_{X}$ is correct to within $1 \%$ is at least 0.95?

What sample size will make it possible to make the same claim if there are just two candidates? And when there are $n$ candidates? (Exclude $n=1$, even though this case is not uncommon in certain political systems.)

Suppose that a sample consisting of $n$ men is used in the survey described in Problem 7.22 to establish the likelihood of being a bathroom singer. Each person is asked to toss a die and answer YES or NO. The probability $\mathbb {P}\left(Y\right)$ of answering YES is estimated by the average number $P_{n} Y$ of YES answers (that is, the number of YES answers divided by $n$ ). Because $\mathbb {P}\left(Y\right)$ is connected with the probability $\mathbb {P}\left(S\right)$ of being a bathroom singer by $\mathbb {P}\left(Y\right)=\frac{1}{6}+\frac{2}{3} \mathbb {P}\left(S\right)$ (see Solution 7.22), the estimate $P_{n} S$ of $\mathbb {P}\left(S\right)$ is then computed from the formula

Estimate the probability that $P_{n} S$ differs from $\mathbb {P}\left(S\right)$ by more than $\varepsilon >0$. Then, find how large a sample is needed to make sure that $P_{n} S$ differs from $\mathbb {P}\left(S\right)$ by no more than 0.01 with probability at least 0.95 ? Compare this to the sample size needed when the survey question is answered directly without tossing a die.

Explain the error in this Newsweek explanation of the margin of error in a public opinion poll:

The margin of error, calculated according to a textbook statistical formula, varies inversely with the sample size. In general, about 500 responses give a possible error of $5 \%$ either way; 2,500 responses decrease it to $1 \%$.

The population of Spain is approximately four times the population of Belgium. If random samples are taken of 2,000 Spaniards and 2,000 Belgians, will the margin for error in these polls be (a) four times larger for the Spanish poll; (b) four times larger for the Belgian poll; (c) twice as large for the Spanish poll; (d) twice as large for the Belgian poll; or (e) the same for both polls?

Does the width of a confidence interval depend on the size of the sample, the size of the population, or the ratio of the sample size to the population size? Explain your reasoning in words not equations.

Americans drive about 3 trillion miles a year and average about 77 reported injuries and 1.09 fatalities per 100 million miles driven. A Google fleet of 55 cars had 11 crashes (2 injuries and no fatalities) while driving 1.3 million miles in autonomous mode between 2009 to 2015.

If a test of autonomous vehicles has no fatalities, how many miles would have to be driven fatality free to reject at the $5 \%$ level the null hypothesis that the probability of a fatality is 1 per 100 million miles?