Fact-checked by Grok 2 weeks ago

Z -test

The Z-test is a fundamental statistical hypothesis test employed to determine whether the mean of a sample is significantly different from a hypothesized population mean, under the condition that the population standard deviation is known.^[1] It relies on the standard normal distribution to evaluate the null hypothesis, typically formulated as H_0: \mu = \mu_0, against alternatives such as one-sided or two-sided deviations.^[2] This test is particularly applicable when the sample size is large (often n \geq 30), leveraging the Central Limit Theorem to approximate the sampling distribution of the mean as normal, even if the underlying population is not perfectly normal.^[1] Key assumptions include the known population variance \sigma^2, random sampling, and independence of observations; violations, such as unknown variance, necessitate alternatives like the t-test.^[2] The test statistic is calculated as Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}, where \bar{x} is the sample mean and n is the sample size, yielding a value that is compared to critical values from the standard normal table or assessed via p-values to decide on rejecting the null hypothesis.^[1] In practice, the Z-test is widely used in fields like quality control, social sciences, and public health for inference on population parameters, such as testing if an observed average IQ score matches a national norm of 100.^[2] Its simplicity and reliance on asymptotic normality make it a cornerstone of parametric statistics, though it assumes ideal conditions that are rarely met exactly in real-world data, prompting robustness checks or non-parametric alternatives when needed.^[1]

Overview

Definition

The Z-test is a parametric statistical hypothesis test designed to assess whether there is a significant difference between a sample statistic and a hypothesized population parameter, typically under the condition that the population variance is known.^[3] It is commonly applied to means or proportions, transforming the observed data into a standardized form to evaluate deviations from the null hypothesis.^[2] The origins of the Z-test trace back to the early 20th century, when Ronald Fisher formalized modern hypothesis testing as part of his contributions to parametric statistical inference, notably in his 1925 book Statistical Methods for Research Workers.^[4] Fisher's work integrated the Z-test into the broader framework of significance testing, emphasizing its role in scientific inference for normally distributed populations.^[5] In contrast to non-parametric tests, which make no assumptions about the underlying data distribution and are thus distribution-free, the Z-test explicitly relies on the normality assumption to derive its probabilistic conclusions.^[6] This reliance on parametric conditions distinguishes it from alternatives like the Wilcoxon signed-rank test, enabling more powerful inferences when the assumptions hold.^[7] The test's foundation in the Z-score concept relates sample data to the standard normal distribution, facilitating straightforward probability calculations for hypothesis evaluation.^[8]

Purpose and Applicability

The Z-test serves as a fundamental tool in statistical hypothesis testing, primarily employed to evaluate claims about population parameters such as means or proportions under conditions where the population variance is known.^[9] It facilitates inference regarding the location parameter (e.g., the population mean μ) in one-sample or two-sample scenarios, or the proportion parameter (e.g., the population proportion p in binomial distributions).^[10] This test is particularly valuable for determining whether observed sample data provide sufficient evidence to reject a null hypothesis about these parameters, enabling decisions in various inferential contexts.^[11] The Z-test is applicable in situations involving large sample sizes, typically n > 30, where the central limit theorem ensures the sampling distribution approximates normality even if the underlying population does not, or when the population is exactly normally distributed regardless of sample size.^[12] It finds widespread use in fields such as quality control, where manufacturers assess whether production processes meet specified mean standards; in surveys, to compare observed proportions against expected population values; and in experimental design, to validate treatment effects on means or proportions derived from controlled trials.^[13]^[10]^[14] These applications leverage the test's reliance on known variance to draw reliable conclusions from sample data about broader populations. A key advantage of the Z-test lies in its enhanced statistical power and precision when the population variance is known, as this eliminates estimation error, resulting in narrower confidence intervals compared to alternatives like the t-test.^[15] This leads to more sensitive detection of true effects, making it preferable in scenarios with ample prior information on variability.^[16]

Assumptions and Conditions

Sampling Assumptions

The Z-test assumes that the sample is drawn as a simple random sample (SRS) from the population, ensuring representativeness and unbiased estimation. Additionally, observations within the sample must be independent, meaning the value of one observation does not influence another. Violations, such as clustered or time-series data, can invalidate the standard normal approximation of the test statistic, leading to erroneous conclusions. These assumptions underpin the reliability of the Central Limit Theorem and the normality of the sampling distribution.^[3]

Normality Requirements

The Z-test relies on the assumption that the population from which the sample is drawn follows a normal distribution, ensuring that the test statistic follows a standard normal distribution under the null hypothesis.^[9] This normality requirement is fundamental for the test's validity, particularly when the population standard deviation is known. However, the Central Limit Theorem (CLT) provides a key relaxation: for sufficiently large sample sizes (typically n > 30), the sampling distribution of the sample mean approximates a normal distribution regardless of the underlying population distribution, allowing the Z-test to be robust to non-normality.^[17] For small samples (n < 30), exact normality of the population is required; deviations from normality can distort the sampling distribution of the test statistic, leading to inflated Type I and Type II error rates and reduced test power.^[9]^[18] In such cases, the Z-test's p-values may no longer accurately reflect the probability of observing the data under the null hypothesis, potentially resulting in incorrect inferences. The role of sample size in invoking the CLT underscores why larger samples mitigate these risks.^[17] To assess the normality assumption, researchers can employ graphical diagnostics such as Q-Q plots, which compare sample quantiles to those expected under a normal distribution, or formal statistical tests like the Shapiro-Wilk test, which evaluates the correlation between observed data and normal scores.^[19] These methods help identify deviations, though they are most reliable for moderate sample sizes and should guide decisions on whether to proceed with the Z-test or opt for alternatives.

Known Variance and Sample Size

The Z-test requires knowledge of the population variance, denoted as σ², to compute the test statistic accurately, as it standardizes the deviation of the sample mean from the hypothesized population mean using the true population standard deviation σ. This parameter is typically derived from prior research, extensive historical data in established processes, or industry standards where variability has been well-characterized over time. When the population variance is unknown, the Z-test is inappropriate, and researchers should instead employ the t-test, which incorporates an estimate of the variance from the sample data to adjust for additional uncertainty.^[3]^[20]^[9] Sample size plays a critical role in ensuring the reliability of the Z-test. A common guideline is that the sample size n should be at least 30, which allows the Central Limit Theorem to approximate the sampling distribution of the mean as normal, even for non-normal populations. Smaller samples are acceptable only if the population distribution is known to be normal and the variance is known, as the exact normal assumption then suffices without relying on asymptotic approximations.^[21]^[9] Violations of these conditions can compromise the test's validity. Substituting the sample standard deviation for the unknown population standard deviation in the Z-test formula leads to biased p-values and inflated Type I error rates, particularly with small samples, as the procedure fails to account for the extra variability in estimating σ and becomes overly sensitive to deviations. Furthermore, inadequate sample sizes diminish the test's statistical power, reducing its ability to detect genuine differences between the sample and population means.^[9]^[12]

Procedure

Formulating Hypotheses

In hypothesis testing using the Z-test, the process begins with formulating the null hypothesis (H_0) and the alternative hypothesis (H_1 or H_a), which represent competing statements about a population parameter. The null hypothesis typically posits no effect or no difference, stating equality between the parameter and a specified value; for testing a population mean, this is expressed as H_0: \mu = \mu_0, where \mu_0 is the hypothesized mean value, while for a population proportion, it is H_0: p = p_0, with p_0 as the hypothesized proportion.^[22]^[23] These formulations assume the null hypothesis is true unless sufficient evidence from the sample data suggests otherwise, serving as the baseline for statistical inference.^[24] The alternative hypothesis specifies the research claim or deviation from the null, which can be two-sided or one-sided depending on the question of interest. A two-sided alternative indicates inequality, such as H_1: \mu \neq \mu_0 for means or H_1: p \neq p_0 for proportions, testing for any difference without direction. One-sided alternatives are directional: H_1: \mu > \mu_0 or H_1: \mu < \mu_0 for means, and similarly H_1: p > p_0 or H_1: p < p_0 for proportions, used when prior knowledge suggests a specific direction of effect.^[22]^[23] The choice between these forms is guided by the study's objectives, ensuring the test aligns with the intended inference.^[24] Associated with these hypotheses are two types of potential errors in decision-making. A Type I error occurs when the null hypothesis is rejected despite being true, representing a false positive, with its probability denoted by \alpha, the significance level—commonly set at 0.05 to control the risk of erroneous rejection.^[25]^[22] Conversely, a Type II error happens when the null hypothesis is not rejected even though it is false, a false negative, with probability \beta, which depends on factors like sample size and the true effect magnitude but is not directly controlled in formulation.^[25]^[22] The significance level \alpha thus defines the threshold for rejecting H_0, balancing the risks of these errors based on the context's consequences.^[26]

Calculating the Test Statistic

The Z-test statistic for a one-sample test of the population mean is calculated using the formula

Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}},

where \bar{x} is the observed sample mean, \mu_0 is the hypothesized population mean under the null hypothesis, \sigma is the known population standard deviation, and n is the sample size.^[3]^[2] This standardization transforms the sample mean into a z-score under the standard normal distribution when the null hypothesis holds.^[27] Once the test statistic Z is computed, the p-value is determined by finding the probability of observing a value at least as extreme as |Z| under the null distribution, using the cumulative distribution function of the standard normal. For a two-tailed test, this is calculated as P(Z > |z_{\text{observed}}|) \times 2, typically via a standard normal table, statistical software, or functions like NORM.S.DIST in Excel.^[2]^[28] The null hypothesis is rejected if the p-value is less than the significance level \alpha.^[29] Alternatively, the critical value approach compares the absolute value of the observed Z to the critical value from the standard normal distribution. For a two-tailed test at \alpha = 0.05, the critical value is z_{\alpha/2} = 1.96; the null hypothesis is rejected if |Z| > 1.96.^[29]^[30] This method defines rejection regions in the tails of the normal distribution corresponding to \alpha.^[31] For testing a single population proportion, the Z-test statistic is given by

Z = \frac{\hat{p} - p_0}{\sqrt{p_0(1 - p_0)/n}},

where \hat{p} is the sample proportion, p_0 is the hypothesized population proportion, and n is the sample size.^[32]^[33] The p-value and critical value procedures follow the same steps as for the mean test, assuming the standard normal approximation holds.^[32]

Applications

Testing Population Means

The one-sample Z-test is employed to assess whether the mean of a sample drawn from a population significantly differs from a specified hypothesized population mean, under the condition that the population standard deviation is known. This test is applicable when the population is normally distributed or the sample size is sufficiently large to invoke the central limit theorem. For instance, in manufacturing, it can evaluate if the average output quality, such as the mean weight of assembled components, aligns with a target specification to ensure process consistency.^[34]^[3] The two-sample Z-test extends this approach to independent samples from two populations, testing the null hypothesis that their population means are equal when the population standard deviations are known. The test statistic is calculated as

Z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}

where \bar{x}_1 and \bar{x}_2 are the sample means, \sigma_1^2 and \sigma_2^2 are the known population variances, and n_1 and n_2 are the sample sizes; this formula accommodates cases of equal or unequal known variances, assuming sample independence and normality or large sample sizes.^[35]^[8] For paired samples, where observations are dependent (e.g., before-and-after measurements on the same subjects), the Z-test is not suitable due to the violation of independence; a paired t-test is recommended instead, as it accounts for the correlation within pairs.^[36] In practice, Z-tests for population means occur frequently in manufacturing to compare mean defect rates between production shifts or machines, helping maintain quality standards. Similarly, in medicine, they are applied to infer mean drug efficacy from continuous outcomes, such as average reduction in blood pressure across treatment groups, when population variability is established from prior studies.^[37]^[38]

Testing Proportions

The Z-test for proportions applies to binomial data, where the goal is to infer about a population proportion p using the normal approximation to the binomial distribution for large samples. This test is particularly suited for categorical outcomes, such as success/failure or yes/no responses, distinguishing it from tests for continuous means. In the one-sample setting, it evaluates whether the population proportion equals a hypothesized value p_0, as in assessing voter preference in surveys where a simple random sample yields the number of "yes" responses out of n trials. The test statistic is calculated as

z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}},

where \hat{p} = x/n is the observed sample proportion and x is the number of successes.^[39] This statistic follows a standard normal distribution under the null hypothesis when the large-sample approximation holds, requiring n p_0 \geq 5 and n(1 - p_0) \geq 5 to ensure the sampling distribution of \hat{p} is approximately normal.^[40] For example, in a survey of 1,000 voters to test if support for a policy is 50% (p_0 = 0.5), with 520 affirmative responses (\hat{p} = 0.52), the conditions are met (n p_0 = 500 \geq 5), and the z-statistic can be computed to determine if the observed proportion significantly differs from 50%.^[41] In the two-sample case, the Z-test compares proportions from two independent binomial populations, such as conversion rates between marketing campaigns or disease prevalence across groups. Under the null hypothesis p_1 = p_2, the test uses a pooled proportion \bar{p} = (x_1 + x_2)/(n_1 + n_2) to estimate the common proportion, yielding the test statistic

z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\bar{p}(1 - \bar{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}},

where \hat{p}_1 = x_1/n_1 and \hat{p}_2 = x_2/n_2.^[42] The approximation requires n_1 \bar{p} \geq 5, n_1(1 - \bar{p}) \geq 5, n_2 \bar{p} \geq 5, and n_2(1 - \bar{p}) \geq 5 for each group to validate normality.^[43] This setup assumes independent samples and is appropriate for unrelated groups, such as comparing website click-through rates from two ad designs in marketing, where one campaign might show 150 conversions out of 1,000 impressions (\hat{p}_1 = 0.15) and another 120 out of 800 (\hat{p}_2 = 0.15), but testing for differences after pooling.^[44] Applications of proportion Z-tests are widespread in fields requiring comparison of binary outcomes. In marketing, they evaluate differences in conversion rates between A/B test variants to optimize campaigns, helping determine if one version significantly outperforms another in user engagement.^[44] In epidemiology, the test assesses disease prevalence across populations, such as comparing cancer rates in dogs exposed versus unexposed to a herbicide (e.g., 191/491 exposed vs. 304/945 unexposed, yielding z \approx 2.58, p-value < 0.01, indicating significant association).^[43] These uses leverage the test's efficiency for large samples, providing p-values to gauge evidence against the null while adhering to the normality conditions for reliable inference.

Examples

One-Sample Mean Test

Consider a scenario in a manufacturing plant where the factory claims that the average weight of produced widgets is \mu = 50 grams, with a known population standard deviation \sigma = 5 grams. A quality control inspector randomly selects a sample of n = 100 widgets and observes a sample mean \bar{x} = 51 grams, prompting a one-sample Z-test to assess whether this deviation from the claimed mean is statistically significant. Following the standard procedure for a one-sample Z-test for the mean, the null hypothesis is H_0: \mu = 50 grams, and the alternative hypothesis is H_1: \mu \neq 50 grams for a two-tailed test. The test statistic is then computed using the formula

z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{51 - 50}{5 / \sqrt{100}} = \frac{1}{0.5} = 2,

where \mu_0 is the hypothesized population mean under H_0. Under the null hypothesis, the Z-statistic follows a standard normal distribution. The corresponding two-tailed p-value is approximately 0.0456, determined from the cumulative distribution function of the standard normal where P(|Z| > 2) = 2 \times (1 - \Phi(2)). At a significance level of \alpha = 0.05, the p-value is less than \alpha, leading to rejection of the null hypothesis. This result indicates statistically significant evidence that the true population mean widget weight differs from 50 grams and likely exceeds it based on the sample. Practically, however, the 1-gram difference may have limited real-world impact if production tolerances allow for such variation, emphasizing the need to evaluate both statistical and practical significance in decision-making. To quantify the magnitude of this difference independent of sample size, the effect size is measured using Cohen's d = Z / \sqrt{n} = 2 / 10 = 0.2, which corresponds to a small effect.^[45]

Two-Proportion Test

A common application of the two-proportion Z-test arises in clinical trials comparing the efficacy of two treatments. Consider a scenario where researchers evaluate two drugs for treating a condition: Drug A is administered to a sample of 100 patients, with 70 successes (success rate \hat{p}_A = 0.7), while Drug B is given to another sample of 100 patients, with 60 successes (success rate \hat{p}_B = 0.6). The null hypothesis is H_0: p_A = p_B, where p_A and p_B are the true population success rates, against the alternative H_a: p_A \neq p_B. To compute the test statistic, first calculate the pooled proportion \bar{p} = \frac{70 + 60}{100 + 100} = 0.65. The standard error under the null hypothesis is \sqrt{\bar{p}(1 - \bar{p}) \left( \frac{1}{n_A} + \frac{1}{n_B} \right)} = \sqrt{0.65 \times 0.35 \times \left( \frac{1}{100} + \frac{1}{100} \right)} \approx 0.0674. The Z-statistic is then Z = \frac{\hat{p}_A - \hat{p}_B}{\text{SE}} = \frac{0.7 - 0.6}{0.0674} \approx 1.48. The two-tailed p-value is approximately 0.139, calculated from the standard normal distribution. Since 0.139 > 0.05, we fail to reject the null hypothesis at \alpha = 0.05. For finite samples, a continuity correction can improve the normal approximation by adjusting the difference in proportions by subtracting $0.5 / n from the absolute difference (Yates' correction), where n is the sample size per group; this modifies the numerator to |\hat{p}_A - \hat{p}_B| - 0.5 / n = 0.1 - 0.005 = 0.095, yielding |Z| ≈ 1.41 and a p-value around 0.159, still leading to failure to reject H_0. This correction, akin to Yates' adjustment in chi-square tests, addresses the discrete nature of binomial data but is less critical for large samples.^[46]^[47] The result indicates no statistically significant difference in success rates between the drugs at the 5% level. For further insight, a 95% confidence interval for the difference p_A - p_B can be constructed as (\hat{p}_A - \hat{p}_B) \pm 1.96 \times \text{SE}, yielding approximately (-0.032, 0.232), which includes zero and aligns with the non-rejection of H_0. This interval provides a range of plausible differences, emphasizing the test's role in inference rather than proof of equality.

References

[1]
10.1 - Z-Test: When Population Variance is Known | STAT 415
It is completely unrealistic to think that we'd find ourselves in the situation of knowing the population variance, but not the population mean.
[2]
[PDF] The Z-test
Jan 9, 2021 · The z-test is a hypothesis test to determine if a single observed mean is significantly different (or greater or less than) the mean under the ...
[3]
8.2.3.3 - One Sample Mean z Test (Optional) | STAT 200
A one sample mean z test is used when the population is known to be normally distributed and when the population standard deviation ( σ ) is known.
[4]
Chapter 9 Hypothesis testing – Introduction to Statistics for Psychology
Fisher was a statistician from London and is noted as the first person to formalize the process of hypothesis testing.
[5]
[PDF] Statistical Hypothesis Tests - Kosuke Imai
Mar 24, 2013 · Ronald Fisher invented the idea of statistical hypothesis testing. He showed, for the first time in the human history, how one can randomize the ...
[6]
SticiGui Approximate Hypothesis Tests: the z Test and the t Test
Jun 17, 2021 · The z test is based on the normal approximation; the t test is based on Student's t curve, which approximates some probability histograms better ...Missing: explanation | Show results with:explanation
[7]
Nonparametric Tests vs. Parametric Tests - Statistics By Jim
Nonparametric tests don't require that your data follow the normal distribution. They're also known as distribution-free tests and can provide benefits in ...
[8]
https://www.stat.ucla.edu/~cochran/stat10/winter/lectures/lect21.html
[9]
Z Test: Uses, Formula & Examples - Statistics By Jim
A Z test compares group means, using samples to determine if a population mean differs from a hypothesized value, or if two population means differ.
[10]
Definition & Two Proportion Z-Test - Statistics How To
A Z-test is a type of hypothesis test—a way for you to figure out if results from a test are valid or repeatable. For example, if someone said they had found a ...
[11]
Z-Test for Statistical Hypothesis Testing Explained | Built In
Oct 16, 2024 · A Z-test is a statistical test to determine if a mean is part of a normal distribution, and if there are significant differences between ...
[12]
T-test vs. Z-test: When to Use Each - DataCamp
Aug 15, 2024 · Use t-tests when dealing with small samples or unknown variance, and Z-tests when samples are large and variance is known.Quick Summary: t-tests vs. Z... · An Introduction to Hypothesis... · Types of Z-tests
[13]
Experimental investigation on Statistical Precision with Z-Tests
Quality control: Manufacturers can use the z-test to determine whether the mean quality of a production process is consistent with a specified standard. Market ...
[14]
Z-7: Hypothesis Testing, Tests of Significance, and Confidence ...
The ideas presented here will be very helpful in making good decisions on the basis of the data collected in an experimental study. EdD, Assistant Professor
[15]
Difference Between Z-Test and T-Test - Analytics Vidhya
Apr 4, 2025 · A z-test is used to test a Null Hypothesis if the population variance is known, or if the sample size is larger than 30, for an unknown ...What is the Z-Test Statistic? · Examples of Z Test · What is the T-Test?
[16]
Key Differences Between Z-Test Vs T-Test - Simplilearn.com
May 10, 2025 · The parametric test assumes that the variables are measured on an interval scale, whereas the non-parametric test assumes that they are measured ...
[17]
7.4 - Central Limit Theorem - STAT ONLINE
The Central Limit Theorem states that if the sample size is sufficiently large then the sampling distribution will be approximately normally distributed.
[18]
[PDF] Non-Normality and Heteroscedasticity in Regression and ANOVA
The effect of non-normality on the Type I and Type II error rates is reduced as the sample size increases. For more information on the experimental design and ...
[19]
Normality Tests for Statistical Analysis: A Guide for Non-Statisticians
According to the available literature, assessing the normality assumption should be taken into account for using parametric statistical tests. It seems that the ...
[20]
8.3: z-Test for a Mean
### Summary of Z-Test Assumptions, Known Population SD, Sample Size, and Why Not Use If Unknown
[21]
Central Limit Theorem | Formula, Definition & Examples - Scribbr
Jul 6, 2022 · By convention, we consider a sample size of 30 to be “sufficiently large.” When n < 30, the central limit theorem doesn't apply. The ...Sample Size And The Central... · Continuous Distribution · Discrete DistributionMissing: Z- unknown
[22]
[PDF] 9 Hypothesis Tests
Statistical hypothesis: a claim about the value of a parameter or population characteristic. Examples: • H: μ = 75 cents, where μ is the true population. ...
[23]
8.1.2 - Hypothesis Testing - STAT ONLINE
The hypothesized value of the population proportion is symbolized by p 0 because this is the value in the null hypothesis ( H 0 ). If ...
[24]
[PDF] Tests of Hypotheses Using Statistics - Williams College
Test at the α = .05 level. To solve this problem, we first need to formulate the null and alternative hypotheses for this test: H0 : µ ≤ 22. H1 : µ > 22. (2.18).
[25]
6: Hypothesis Testing, Part 2 - STAT ONLINE
Identify Type I and Type II errors; Select an appropriate significance level (i.e., α level) for a given scenario; Explain the problems associated with ...
[26]
Type I and II Errors and Significance Levels
May 12, 2011 · A Type I error would correspond to convicting an innocent person; a Type II error would correspond to setting a guilty person free.
[27]
Hypothesis Testing - Data Science Discovery
You can do a z-test for means or for proportions. This is the most simple type of hypothesis test and it uses z-scores and the normal curve. Let's look at one ...
[28]
Data analysis: hypothesis testing: 6. 1 Calculating the p-value
To find the p-value corresponding to the z-statistic, we use Excel's NORM.S.DIST(z, cumulative) function. The calculation method depends on the specific null ...<|control11|><|separator|>
[29]
S.3.1 Hypothesis Testing (Critical Value Approach) - STAT ONLINE
Specify the null and alternative hypotheses. Using the sample data and assuming the null hypothesis is true, calculate the value of the test statistic.
[30]
[PDF] 9.2 Critical Values for Statistical Significance in Hypothesis testing
A sample mean with a z-score greater than or equal to the critical value of 1.645 is significant at the 0.05 level. There is 0.05 to the right of the critical ...
[31]
Chapter 10: Hypothesis Testing with Z - Maricopa Open Digital Press
Using z is an occasion in which the null hypothesis is a value other than 0. For example, if we are working with mothers in the U.S. whose children are at risk ...
[32]
S.6 Test of Proportion | STAT ONLINE - Penn State
State the null hypothesis H0 and the alternative hypothesis HA. · Calculate the test statistic: z = p ^ − p 0 p 0 ( 1 − p 0 ) n · Determine the critical region.
[33]
7.4 Inference for a Proportion – Significant Statistics
For the normal distribution of proportions, the z-score formula is as follows: If \hat{p} ~N \left(p,\sqrt{\frac{p\cdot q} then the z-score formula is: z ...
[34]
7.2.2. Are the data consistent with the assumed process mean?
If the standard deviation is known, the form of the test statistic is z = Y ¯ − μ 0 σ / N . For case (1), the test statistic is compared with ...
[35]
[PDF] Key Formulas - From Larson/Farber Elementary Statistics
Two-Sample z-Test for the Difference Between Means. (σ and σ2 are known, the samples are random and independent, and either the populations are normally.
[36]
The Differences and Similarities Between Two-Sample T-Test ... - NIH
Two-sample t-test is used when the data of two samples are statistically independent, while the paired t-test is used when data is in the form of matched pairs.
[37]
7.3.1. Do two processes have the same mean?
The question being addressed is whether the mean, μ 2 , of the new assembly process is smaller than the mean, μ 1 , for the old assembly process (process 1).
[38]
Sample Size Estimation in Clinical Trial - PMC - NIH
The aim of this article is to discuss how important sample size estimation is for a clinical trial, and also to understand the effects of sample size over- ...
[39]
8.1 - One Sample Proportion | STAT 200
According to the Rule of Sample Proportions, if n p ≥ 10 and n ( 1 − p ) ≥ 10 then the sampling distributing will be approximately normal. When constructing a ...
[40]
[PPT] Lecture note
Under certain conditions, [np > 5 and n(1-p) > 5], is approximately normally distributed, with m = p and s2 = p(1 - p)/n.
[41]
(PDF) Estimation of a Proportion with Survey Data - ResearchGate
Aug 10, 2025 · Estimating the proportion of voters in favour of a political party, based on a political opinion survey, is just one concrete example of this ...<|separator|>
[42]
9.1 - Two Independent Proportions - STAT ONLINE
Two independent proportions tests are used to compare the proportions in two unrelated groups. In StatKey these were known as "Difference in Proportions" tests.Missing: applications | Show results with:applications
[43]
6.2 Difference of two proportions
The difference of two proportions (p1-p2) is estimated by ^p1−^p2. Its standard deviation is calculated as √p1(1−p1)n1+p2(1−p2)n2.
[44]
[PDF] Two Sample Z Test For Proportions
two sample z test for proportions is a fundamental statistical tool used to compare the proportions of two independent groups. Whether you're analyzing ...
[45]
[PDF] Two tails of Z
Apr 7, 2007 · Two tails of Z. Entries in the table represent two-tailed P values for z statistics hundredths tenths. 0.00. 0.01. 0.02. 0.03. 0.04. 0.05. 0.06.
[46]
One-Sample Z Test – Victor Bissonnette
Cohen's D: (.36): Cohen's D is an effect-size estimate. It is equal to the mean difference divided by the standard deviation of the population. It standardizes ...
[47]
9.4 - Comparing Two Proportions | STAT 415
The test statistic for testing the difference in two population proportions, that is, for testing the null hypothesis ... 10.1 - Z-Test: When Population Variance ...
[48]
[PDF] Inference for Proportions
Oct 17, 2020 · We can test null hypotheses about π by comparing the observed Z to the quantiles of a standard normal distribution. Yate's correction for ...
[49]
6a.7 - Example: Comparative Treatment Efficacy Studies
... z-test using a normal approximation, a χ 2 test (basically, a square of the z-test), a χ 2 test with continuity correction, and Fisher's exact test.