Example 7.1.1 At a light bulb factory, the average life expectancy of a light bulb made by a conventional production method is known to be 1500 hours and the standard deviation is 200 hours. Recently, the company is trying to introduce a new production method, with the average life expectancy of 1600 hours for light bulbs. To confirm this argument, 30 samples were taken from the new type of light bulbs by simple random sampling and the sample mean was \(\small \overline x \) = 1555 hours. Can you tell me that the new type of light bulb has the average life of 1600 hours?

Answer

A statistical approach to the question of this issue is first to make two assumptions about the different arguments for the population mean μ . Namely, $$ \small \begin{multline} \shoveleft H_0 : μ = 1500 \\ \shoveleft H_1 : μ = 1600 \end{multline} $$ \(\small H_0\) is called a null hypothesis and \(\small H_1\) is an alternative hypothesis. In most cases, the null hypothesis is defined as an ‘existing known fact’, and the alternative hypothesis is defined as ‘new facts or changes in current beliefs’. So when choosing between two hypotheses, the basic idea of testing hypothesis is 'unless there is a significant reason, we accept the null hypothesis (current fact) without choosing the alternative hypothesis (the fact of the matter). This idea of testing hypothesis is referred to as a ‘conservative decision making’.

A common sense criterion for choosing between two hypotheses would be 'which population mean of two hypothesis is closer in distance to the sample mean'. Based on this common sense criteria which uses the concept of distance, the sample mean of 1555 is closer to \(\small H_1 : μ = 1600\) so the alternative hypothesis will be chosen. A statistical testing hypothesis is based not only on this common sense criteria, but also on the sampling distribution of \(\small \overline X\). In other words, the statistical testing hypothesis is to select a critical value \(C\) based on the sampling distribution theory and to make a decision rule as follows:

\( \small \qquad \text { ‘If \(\overline X\) is smaller than C, then the null hypothesis \(H_0\) will be chosen, else reject \(H_0\)’} \)

The area of {\(\small \overline X < C\)} is called an acceptance region of \(\small H_0\) and the area {\(\small \overline X ≥ C\)} is called a rejection region of \(\small H_0\) (<Figure 7.1.1>).

If a hypothesis is chosen by this decision rule, there are always two possible errors in the decision. One is a Type 1 Error which accepts \(\small H_1\) when \(\small H_0\) is true, the other is a Type 2 Error which accept \(\small H_0\) when \(\small H_1\) is true. These errors can be summarized as Table 7.1.1.

Table 7.1.1 Two types of errors in testing hypothesis

	Actual \(\small H_0\) is true	Actual \(\small H_1\) is true
Decision : \(\small H_0\) is true	Correct	Type 2 Error
Decision : \(\small H_1\) is true	Type 1 Error	Correct

If you try to reduce one type of error when the sample size is fixed, then the other type of error is increasing. That is why we came up with a conservative decision making method that defines the null hypothesis \(\small H_0\) as 'past or present facts' and 'accept the null hypothesis unless there is a significance evidence for the alternative hypothesis.' In this conservative way, we try to reduce the type 1 error as much as possible that selects \(\small H_1\) when \(\small H_0\) is true, which would be more risky than the type 2 error. Testing hypothesis determines the tolerance for the probability of the type 1 error, usually 5% or 1% for rigorous test, and use the selection criteria that satisfy this limitation. The tolerance for the probability that this type 1 error will occur is called the significance level and is often expressed as α. The probability of the type 2 error is expressed as β.

If the significance level is established, the decision rule for the two hypotheses can be tested using the sampling distribution of all possible sample means in Chapter 6. <Figure 7.1.2> shows the distribution of populations for two hypotheses, and the distribution of all possible sample means for each population.

If the population corresponds to the distribution of \(\small H_0\) : μ = 1500, the sampling distribution of all possible sample means is approximated as \(\small N(1500,200^2 )\) by the central limit theorem. If the population corresponds to the distribution of \(\small H_1\) : μ = 1600, the sampling distribution of all possible sample means is approximated as \(\small N(1600,200^2 )\). The standard deviation for each population is assumed to be 200 from a historical data. Then the decision rule becomes as follows:

\( \small \qquad \text {‘If \(\overline X < C\), then accept \(H_{0}\), else accept \(H_{1}\) (i.e. reject \(H_{0}\) )’} \)

In Figure 7.1.2, the shaded area represents the probability of the type 1 error. If we set the significance level, which is the tolerance level of the type 1 error, is 5%, i.e. \(\small P(\overline X < C) = 0.95\), \(C\) can be calculated by finding the percentile of the normal distribution \(\small N(1500,\frac{200^2}{30})\) as follows:

\( \small \qquad 1500+1.645 \frac{200}{\sqrt 30 } = 1560.06 \)

Therefore, the decision rule can be written as follows.

\( \small \qquad \text {‘If \(\overline X\) < 1560.06, then accept \(H_0 \), else reject \(H_0\) (accept \(H_1\) ).’} \)

In this problem, the observed sample mean of the random variable \(\small \overline X\) is \(\small \overline x\)= 1555 and \(\small H_0 \) is accepted. In other words, the hypothesis of \(\small H_0 \) : μ = 1500 is judged to be correct, which contradicts the result of common sense criteria that \(\small \overline x\) = 1555 is closer to \(\small H_1 \) : μ = 1600 than \(\small H_0 \) : μ = 1500. This result can be interpreted that the sample mean of 1555 is not a sufficient evidence to reject the null hypothesis by a conservative decision making method.

The above decision rule is often written as follows, emphasizing that it is the result from a conservative decision making method.

\( \small \qquad \text {‘If \(\overline X\) < 1560.06, then do not reject \(H_0\), else reject \(H_0 \).’} \)

In addition, this decision rule can be written for calculation purpose as follows.

\( \small \qquad \text {‘If \(\frac{\overline X - 1500}{\frac {200}{\sqrt{30}}}\), then accept \(H_0\) , else reject \(H_0\).’} \)

In this case, since \(\small\overline x\) = 1555, \(\frac{1555 - 1500}{\frac {200}{\sqrt{30}}}\) and it is less than 1.645. Therefore, we accept \(\small H_0\).

Example 7.1.2 The weight of a bag of cookies is supposed to be 250 grams. Suppose the weight of all bags of cookies is a normal distribution. In the survey of 100 samples of bags which were randomly selected, the sample mean was 253 grams and the standard deviation was 10 grams.

Answer

$$ \small \begin{multline} \shoveleft \text{'If } \frac {\overline X - \mu_0}{ \frac {S}{\sqrt{n}} } > z_{α} , \text{ then reject } H_0 \text{ else accept } H_0 ’ \\ \shoveleft \text{'If } \frac {253 - 250}{ \frac {10}{\sqrt{100}} } > z_{0.01} , \text{ then reject } H_0 \text{ else accept } H_0 ’ \\ \end{multline} $$ $$ \small \begin{multline} \shoveleft \text{'If } \overline X > \mu_0 + z_{α} \frac {S}{\sqrt{n}} , \text{ then reject } H_0 \text{ else accept } H_0 ’ \\ \shoveleft \text{'If } \overline X > 250 + 2.326 \frac {10}{\sqrt{100}} , \text{ then reject } H_0 \text{ else accept } H_0 ’ \\ \shoveleft \text{'If } \overline X > 252.326 , \text{ then reject} H_0 \text{ else accept } H_0 ’ \end{multline} $$

\( \small \qquad \qquad p\text{-value} = P( \overline X \gt 253) = P( Z \gt \frac{253-250}{\frac{10}{10}} ) = P( Z \gt 3) = 0.0013 \)

$$ \small \begin{multline} \shoveleft \text{'If } \left| \frac {\overline X - \mu_0}{ \frac {S}{\sqrt{n}} } \right| > z_{α/2} , \text{ then reject } H_0 \text{ else accept } H_0 ’ \\ \shoveleft \text{'If } \left| \frac {253 - 250}{ \frac {10}{\sqrt{100}} } \right| > z_{0.005} , \text{ then reject } H_0 \text{ else accept } H_0 ’ \\ \end{multline} $$

\( \small \qquad \qquad p\text{-value} = 2 P(\overline X \gt 253) = 2 P(\overline X \gt \frac{253-250}{\frac{10}{10}} = 2 P( Z \gt 3) = 0.0026 \)

[Testing Hypothesis : Population Mean μ ]

Example 7.1.3 When the sample size is 16 and the sample variance is 100 in [Example 7.1.2], test whether the average weight of the cookie bags is 250g or greater and obtain the \(p\)-value. Check the result using 『eStatU』

Answer

Since the population standard deviation is unknown and the sample size is small, the decision rule is as follows. $$ \small \begin{multline} \shoveleft \text{'If } \frac {\overline X - \mu_0} {\frac {S}{\sqrt{n}} } > t_{n-1: α} , \text{ then reject } H_0 \text{ else accept } H_0 ’ \\ \shoveleft \text{'If } \frac {253 - 250}{ \frac {10}{\sqrt{16}} } > t_{16: 0.01} , \text{ then reject } H_0 \text{ else accept } H_0 ’ \\ \end{multline} $$ Since the value of test statistic is \( \frac {253 - 250}{ \frac {10}{\sqrt{16}} } = 1.2 \), and \(t_{15: 0.01} = 2.602\) , we accept \(\small H_0\). Note that the decision rule can be written as follows.

\( \small \qquad \qquad\text{'If } \overline X > 250 + 2.602 \frac {10}{\sqrt{16}} , \text{ then reject } H_0 \text{ else accept } H_0 ’ \\ \)

In <Figure 7.1.3> of 『eStatU』 , select the right-sided test of [Hypothesis], select the \(t\)-test on [Test Type] and enter sample size \(n\) = 16, then the test result is as <Figure 7.1.6> if you click the [Execute] button.

Since the \(p\)-value is the probability that \(t_{15}\) is greater than the test statistics 1.200, the \(p\)-value is 0.124 by using the module of \(t\) distribution in 『eStatU』.

Example 7.1.4 (Heights of college students)

10 male students are sampled in a university and examined their heights as follows:

Test the hypothesis whether the population mean is 175cm or greater with the significance level of 5%.

Answer

After entering data on a sheet as shown in <Figure 7.1.7> in 『eStat』 , clicking the testing hypothesis for the population mean and then clicking variable name V1 for 'Analysis Var' in variable selection box will result in a dot graph with 95% confidence interval as in <Figure 7.1.8>.

If you click [Histogram] button from option menu below the graph as in <Figure 7.1.9>, the corresponding histogram is appeared as in <Figure 7.1.10>. The histogram together with the normal distribution graph can be used to check whether the sample data comes from a normal distribution. The options such as [Normal Q-Q Plot] and [Normality Test] will be explained in chapter 11.

Enter \(μ_0 \) = 175 at the box of option menu, select the right sided test and the significance level of 5%. Then press the [t-Test] button to display the hypothesis test graph as shown in <Figure 7.1.11> and a test result in the Log Area as in <Figure 7.1.12>.

You can select Z-test in the option menu, but you have to enter the population standard deviation σ in this case.

Example 7.2.1 One company produces bolts for an automobile. If the average diameter of bolts is 15mm and its variance is less than or equal to \(0.1^2\), it can be delivered to the automobile company. Twenty-five of the most recent products were randomly sampled and their variance was \(0.15^2\). Assuming that the diameter of a bolt follows a normal distribution,

Answer

\( \small \qquad \qquad '\text{If } \frac{(n-1)S^2}{\sigma^2_{0}} \gt \chi^2_{n-1: α} , \text{ then reject } H_0 , \text{ else accept } H_0 ’ \)

[]

Example 7.2.2 (Heights of college students)
Using 『eStat』 and the height data of 10 male college students in [Example 7.1.4],

test the hypothesis whether the population variance is greater than 25 at the significance level of 5%.

Answer

After entering data as shown in <Figure 7.2.3> on the sheet in 『eStat』 , click the icon of testing hypothesis for variance and select ‘Height’ as the Analysis Var to display a dot graph of data with (average) ± (standard deviation) interval as in<Figure 7.2.4>.

In the option box under the Graph Area (<Figure 7.2.5>), enter \(\sigma^2_{0}\) = 25, and select the alternative hypothesis as right-sided test, significance level as 5%. By clicking [\(\chi^2\) test] button, the result of the testing hypothesis will be shown as <Figure 7.2.6> and the result table as <Figure 7.2.7>.

Example 7.3.1 A survey was conducted last month for the election of a national assembly member. According to the survey of the last month, the approval rating of a particular candidate was 60 percent. In order to see if there is a change in the approval rating, a sample survey of 100 people has been conducted and 55 people supported it.

Answer

\( \small \qquad \qquad'\text{If } \left | \frac {\hat p - p_0}{ \sqrt { \frac {p_0 (1-p_0)}{n} } } \right | > z_{α/2} , \text{ then reject } H_0 , \text{ else accept } H_0 ' \)

\( \small \qquad \qquad \left | \frac {0.55 - 0.6}{ \sqrt { \frac {0.6 (1-0.6)}{100} } } \right | = 1.005 \)

[Testing Hypothesis : Population Proportion p]

Example 7.4.1 For the testing hypothesis of [Example 7.1.1], calculate the probability of the type 2 error β if the significance level is 5%. Check this result using 『eStatU』.

Answer

The hypothesis in [Example 7.1.1] is \(\small H_0 : \mu = 1500, H_1 : \mu = 1600\), the population standard deviation is assumed σ = 200, the sample size is \(n\) = 30 and hence the decision rule is as follows if the significance level is 5%.

\( \small \qquad '\text{If } \overline X \lt 1500 + 1.645 \times \frac {200}{\sqrt{30}} = 1560.06, \text{ reject } H_0,\text{ else accept } H_0’ \)

Hence the type 2 error which is the probability of '\(\small H_0\) is true when \(\small H_1\) is true’ can be calculated as follows:

\( \small \qquad \beta = P(\overline X \lt 1560.06 | H_1 \) is true)
\( \small \qquad \;\; = P(\frac {\overline X - 1600}{\frac{200}{\sqrt{30}} } < \frac {1560.06 - 1600}{\frac{200}{\sqrt{30}} } ) \)
\( \small \qquad \;\; = P( Z \lt -1.09 ) \)
\( \small \qquad \;\; = 0.137 \)

Select [Testing \( \mu - C, \beta \)] at 『eStatU』 menu. Enter \(\small \mu_0 = 1500 , \mu_1 = 1600, \sigma = 200, \alpha = 0.05, n = 30 \) in the input box window as <Figure 7.4.1> and click the [Execute] button. The result of the testing hypothesis, the critical value \(C\) and the probability of the type 2 error β, will be shown as in <Figure 7.4.2>.

[Testing Hypothesis : Population Mean μ ]

Example 7.4.2 In [Example 7.1.1], if the null hypothesis is not changed, but the alternative hypothesis is changed as follows:

\(\qquad \small H_0 : \mu = 1500 ,\;\; H_1 : \mu = 1580\)

Answer

\(\qquad\)‘If \(\small \overline X \lt 1500 + 1.645 \times \frac {200}{30} = 1560.06\), reject \(\small H_0\), else accept \(\small H_0\)’

Hence the probability of type 2 error is as follows:

\(\qquad \small \beta = P(\overline X \lt 1560.06 \;|\; H_1 \) \( \text{is true} )\)
\(\qquad \small \;\;= P(\frac {\overline X - 1580}{\frac{200}{\sqrt{30}} } < \frac {1560.06 - 1580}{\frac{200}{\sqrt{30}} } ) \)
\(\qquad \small \;\;= P( Z \lt -0.546 ) = 0.283 \)

Example 7.4.3 In [Example 7.1.1], calculate the power of the following alternative hypothesis. Use = 0.05. By using this, approximate the power function.

Answer

Although the alternative hypotheses are different, the decision rule is the same as follows:

\( \qquad \small '\text{If } \overline X \lt 1500 + 1.645 \times \frac {200}{30} = 1560.06, \text{ reject } H_0,\text{ else accept } H_0’\)

Hence if we calculate the probability of the type 2 error as [Example 7.4.2], the power of each test is as follows:

Alternative Hypothesis	β	Power = 1 - β
1) \(\small H_1 : \mu = 1500 \) 2) \(\small H_1 : \mu = 1510 \) 3) \(\small H_1 : \mu = 1520 \) 4) \(\small H_1 : \mu = 1530 \) 5) \(\small H_1 : \mu = 1540 \) 6) \(\small H_1 : \mu = 1550 \) 7) \(\small H_1 : \mu = 1560 \) 8) \(\small H_1 : \mu = 1570 \) 9) \(\small H_1 : \mu = 1580 \) 10) \(\small H_1 : \mu = 1590 \) 11) \(\small H_1 : \mu = 1600 \) 12) \(\small H_1 : \mu = 1610 \)	0.95 0.91 0.86 0.79 0.71 0.61 0.50 0.39 0.29 0.21 0.14 0.09	0.05 0.09 0.14 0.21 0.29 0.39 0.50 0.61 0.71 0.79 0.86 0.91

The power function can be approximated by connecting points of (\(\mu\), 1 - \(\beta\) ) in each test as <Figure 7.4.3>.

Example 7.4.4 Consider the testing hypothesis on the bulb life such as \(\small H_0 : \mu = 1500, H_1 : \mu = 1570 \). Find the sample size \(n\) and the decision rule which satisfies α of 5% and β of 10%. Assume that the population standard deviation σ is 200 hours. Check the result using 『eStatU』.

Answer

Let \(n\) be the sample size and \(\small C\) be the critical value of a decision rule. The probability of the type 1 error α and the probability of the type 2 error β are defined as follows: $$ \small \begin{multline} \shoveleft \alpha = P(\overline X \gt C | H_0 \text{ is true } ) \\ \shoveleft \beta = P(\overline X \lt C | H_1 \text{ is true } ) \\ \end{multline} $$ If \(\small H_0\) is true, the sampling distribution of \(\small \overline X\) is \(\small N(1500, \frac{200^2}{n})\) and if \(\small H_1\) is true, the sampling distribution of \(\small \overline X\) is \(\small N(1570, \frac{200^2}{n})\). If α = 0.05 and β = 0.10, then \(z_{0.05}\) = 1.645 and \(z_{0.90}\) = -1.280. Hence \(n\) and \(\small C\) should satisfy both of the following equations and they can be calculated by solving the two system of equations. $$ \small \begin{multline} \shoveleft C = 1500 + 1.645 \times \frac{200}{\sqrt{n}} \\ \shoveleft C = 1570 - 1.280 \times \frac{200}{\sqrt{n}} \\ \end{multline} $$ The solution is \(n\) = 69.9, \(\small C\) = 1539.4. i.e., the sample size is 70 approximately and the decision rule is as follows: $$ \small \begin{multline} \shoveleft '\text{If } \overline X \gt 1539.4, \text { then reject } H_0, \text { else accept } H_0’ \end{multline} $$ Select [Testing μ - \(C, n\)] at 『eStatU』 menu. Enter \(\mu_0 = 1500 ,\mu_1 = 1570 , \sigma = 200, \alpha = 0.05 \), \( \beta = 0.10 \) in the input box window as <Figure 7.4.4>.

If you click the [Execute] button, you will see the test result of 『eStatU』 as in <Figure 7.4.5>. The critical value and the sample size are calculated.

[Testing Hypothesis : Population Mean μ

Example 7.5.1 (Case of lower specification limit \(L\))

Assume that the tension of the wire string produced by a steel company is normally distributed based on past experience and the variance is 30, that is, the standard deviation is \(\sigma = \sqrt{30} = 5.48 \). If the wire tension of one roll is less than 87, it cannot be used for the desired work and is judged as a defective product. That is, the lower specification limit is \(L\) = 87. Inspecting an entire lot containing many rolls of wire rope is impossible because the products can be destroyed. Therefore, a sample of rolls is selected from the lot, and if the defective rate of the sample is estimated to be 1%, the lot is accepted, and if the defective rate is estimated to be 5%, the lot is rejected.

What should be the sample size and decision criteria? However, note that the decision criteria should satisfy the producer risk probability of rejecting a good lot, \(\alpha\) = 1%, and the consumer risk probability of rejecting a bad lot, \(\beta\) = 10%.

Answer

Assume that the quality characteristic value is the random variable X which follows a normal distribution with the mean of \(\mu_0\) for an accepted lot and the variance of \(\sigma^2\) is 30. Since the lower specification limit is \(L\) = 87, the probability that the characteristic value is less than 87 is 1% can be represented as P(X < 87) = 0.01. Using the standardization of the normal random variable X, \(Z\), it can also be written as follows. $$ P(Z < \frac {87 - \mu_0 } { \sqrt{30} } ) = 0.01 $$ Therefore, according to the standard normal distribution table, \(\frac {87 - \mu_0 } { \sqrt{30} } \) should be –2.33. That is, \(\mu_0 = 87 + 2.33 \sqrt{30} = 99.7684 \) which is the population mean of an accepted lot.

In a similar way, the characteristic values of a rejected lot follows a normal distribution with the mean of \(\mu_1\) and the variance of 30. The probability that the characteristic value is less than 87 is 5% can be represented as P(X < 87) = 0.05. Using the standardization of the normal random variable X, \(Z\), it can also be written as follows. $$ P(Z < \frac {87 - \mu_1 } { \sqrt{30} } ) = 0.05 $$ Therefore, \(\frac {87 - \mu_1 } { \sqrt{30} } \) should be –1.645 and \(\mu_1\)= 96.0146 which is the population mean of a rejected lot.

The problem of selecting a sample from a lot to decide whether accepting the lot or rejecting the lot is the same as the testing hypotheses about the two population means as follows. $$ H_0 : \mu = 99.7684 , H_1 : \mu = 96.0146 $$ For this testing hypothesis, the sample size and decision criteria are determined by using the probability of type 1 error (rejecting a good lot: producer risk), \(\alpha\), and the probability of type 2 error (accepting a bad lot: consumer risk), \(\beta\). (Refer to Section 7.4 ‘Hypothesis test with \(\alpha\) and \(\beta\) simultaneously’).

This hypothesis test is based on the distribution of sample means in Chapter 5. If the population is a normal distribution with a mean of 99.7684 and a variance of 30, the distribution of the sample mean of size \(n\) is \(N(99.7684, 30/n)\), and if the population is a normal distribution with a mean of 96.0146 and a variance of 30, the distribution of the sample mean is \(N(96.0146, 30/n)\). Therefore, the problem is to determine the decision criteria and the sample size based on the probability of rejecting a good lot as \(\alpha\) = 1% and the probability of accepting a bad lot as \(\beta\) = 10%, as shown in <Figure 7.5.1>, based on the distribution of the sample mean. The goal is to find \(C\) and \(n\).

\(\alpha\) = 1% means that the probability that a sample mean \(\overline X\) is smaller than \(C\) in a normal distribution \(N(99.7684, 30/n)\) as shown in <Figure 7.5.1>, that is, \(P(\overline X < C) = 0.01\). Using the standardization of the normal random variable X, it can also be written as follows. $$ P(Z < \frac {C - 99.7684} {\sqrt { \frac{30}{n}} } ) = 0.01 $$ Therefore, based on the standard normal distribution table, it satisfies the following equation. $$ \frac {C - 99.7684} {\sqrt { \frac{30}{n}} } = -2.33 $$ Similarly, \(\beta\) = 10% means that a sample mean \(\overline X\) is larger than \(C\) in a normal distribution \( N(96.0146, 30/n) \), that is, \(P(\overline X > C) = 0.10 \). Using the standardization of the normal random variable X, it can also be written as follows. $$ P(Z > \frac {C - 96.0146} {\sqrt { \frac{30}{n}} } ) = 0.10 $$ Therefore, based on the standard normal distribution table, it satisfies the following equation. $$ \frac {C - 96.0146} {\sqrt { \frac{30}{n}} } = 1.28 $$ If we solve the above two equations with respective to \(C\) and \(n\), the sample size is, \(n\) = 27.77, critical value is, \(C\) = 97.3457. Hence, the decision rule becomes as follows. Select a sample of size 28 and calculate its sample mean and $$ \text {‘If the sample mean} \overline X < 97.3457, \text {then reject the lot, else accept the lot.’} $$ [Testing Hypothesis : Population Mean μ

7.5.2 Acceptance Sampling by Variable: Case of Unknown Lot \(\sigma\)

When the standard deviation of quality characteristic value of a lot, \(\sigma\), is not known, the sample inspection method is essentially the same as when \(\sigma\) is known, except that an estimate of \(\sigma\) is used. If the standard deviation \(\sigma\) is not known, samples are commonly inspected by using Part B in the U.S. Department of Defense Military Standard Sampling Procedures and Tables for Inspection by Variables (MIL-STD-414). This table assumes that the quality characteristic values are normally distributed. There are four parts as follows, and the inspection procedure is explained in each part.

Part A: General information regarding sample inspection.
Part B: Sample inspection, standard deviation method when the standard deviation of the lot is unknown
Part C: Sample inspection, range method when the standard deviation of the lot is unknown
Part D: Sample inspection when the standard deviation of the lot is known

There are five inspection levels of MIL-STD-414, I, II, III, IV, and V, depending on the price of the product or the time required for inspection. The general inspection level is IV. MIL-STD-414 specifies the sample size and the maximum allowable defective rate M of the sample for the lot to pass, given the lot size, quality acceptance level of the entire lot, and inspection level.

The detailed information of MIL-STD-414 is so extensive and it is omitted in this book. Anyone interested should refer to the references.

7.6 Exercise