In a certain hypothesis test, H 0 : p = 0.4 H_0:p=0.4 , H a : p H_{a}:p < 0.4 0.4 . You collect a sample and calculate a test statistic z = − 1.32 z=-1.32 . Find the P P -value .

P ( z P\left(z\right. P ( z < − 1.32 ) = 0.0934 -1.32)=0.0934

P ( z P\left(z\right. P ( z < − 1.32 ) = 0.8132 -1.32)=0.8132

P ( z P\left(z\right. P ( z < − 1.32 ) = 0.9066 -1.32)=0.9066

P ( z P\left(z\right. < − 1.32 ) = 0.1868 -1.32)=0.1868

9. Hypothesis Testing for One Sample

Steps in Hypothesis Testing

9. Hypothesis Testing for One Sample

Steps in Hypothesis Testing: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

In hypothesis testing, the null hypothesis (H₀) represents a claim about a population, while the alternative hypothesis (H_a) challenges this claim. The process involves calculating a test statistic, such as a z-score, to determine how unusual the sample data is compared to the null hypothesis. The p-value, a probability derived from the test statistic, indicates the likelihood of observing the sample data if H₀ is true. A comparison of the p-value to a significance level (α) guides the decision to reject or fail to reject H₀, ultimately informing conclusions about the population parameter.

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Step 1: Write Hypotheses

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Step 1: Write Hypotheses Video Summary

In hypothesis testing, the first crucial step is formulating two statements: the null hypothesis and the alternative hypothesis. These statements are derived from the problem context and are essential for guiding the analysis. The null hypothesis, denoted as $ H_0 $ (or $ H_{\text{naught}} $), represents a claim about a population parameter that we assume to be true. This parameter can be a mean ($ \mu $), proportion ($ p $), or standard deviation ($ \sigma $), and it is typically expressed with an equality, such as $ \mu = 23 $ or $ p = 0.30 $.

For instance, if a researcher investigates the average age of students at a university and the enrollment office claims that the mean age is 23, the null hypothesis would be $ H_0: \mu = 23 $. This statement serves as the default assumption that will be tested against evidence from the data.

The alternative hypothesis, denoted as $ H_a $, represents the opposing claim that the researcher aims to support. It is formulated using the same parameter and value as the null hypothesis but incorporates a different relational symbol: either less than, greater than, or not equal to. In the previous example, if the researcher wants to test whether students are younger than the claimed average, the alternative hypothesis would be $ H_a: \mu < 23 $.

Keywords in the problem statement, such as "greater than," "less than," or "not equal to," help determine the appropriate relational symbol for the alternative hypothesis. For example, if a business journal seeks to estimate the percentage of companies with female CEOs and aims to prove that this percentage is greater than 20%, the null hypothesis would be $ H_0: p = 0.20 $, while the alternative hypothesis would be $ H_a: p > 0.20 $.

Understanding how to construct these hypotheses is vital, as the null hypothesis provides a baseline for testing, while the alternative hypothesis indicates the direction of the test. This foundational knowledge sets the stage for further statistical analysis and decision-making based on the evidence collected.

Problem

A popular theme park claims that their weekly attendance is around $100 comma 000$ . You believe that the weekly attendance is different than this claimed value, so you gather sample data. Write the null and alternative hypotheses.

$H_0:$ $mu equals 100 comma 000$

$H_{a}:$ $mu is greater than 100 comma 000$

$H_0:$ $mu equals 100 comma 000$

$H_{a}:$ $mu is not equal to 100 comma 000$

$H_0:$ $\mu=100,000$

$H_{a}:$ $μ$ < $100 comma 000$

$H_0:$ $\mu\ne100,000$

$H_{a}:$ $\mu=100,000$

Problem

A candy manufacturer seeking to minimize the variation in weights of their candies claims to produce candies with a standard deviation less than $0.300$ g. Write the null and alternative hypotheses.

$H_0:$ $sigma equals 0.300$ g

$H_{a}:$ $sigma is greater than 0.300$ g

$H_0:$ $\sigma=0.300$ g

$H_{a}:$ $sigma is not equal to 0.300$ g

$H_0:$ $\sigma=0.300$ g

$H_{a}:$ $σ$ < $0.300$ g

$H_0:$ $\sigma>0.300$ g

$H_{a}:$ $\sigma$ < $0.300$ g

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Step 1: Write Hypotheses Example 1

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Step 1: Write Hypotheses Example 1 Video Summary

In hypothesis testing, formulating the null and alternative hypotheses is crucial for statistical analysis. In this scenario, we are given a claim regarding the parameter μ (mu), specifically that μ is less than 120. This claim serves as a foundation for our hypotheses.

The null hypothesis, denoted as H₀, represents the default assumption that there is no effect or no difference. In this case, since the claim involves a specific value, we establish the null hypothesis as:

H₀: μ = 120

This indicates that we assume the mean μ is equal to 120 until evidence suggests otherwise. Conversely, the alternative hypothesis, denoted as H_A, reflects the claim we are testing against the null hypothesis. Here, the alternative hypothesis is:

H_A: μ < 120

This hypothesis suggests that the mean μ is indeed less than 120, which is the claim we are seeking to provide evidence for. It is important to note that the alternative hypothesis is often what researchers aim to support through their statistical tests.

In summary, when faced with a claim about a parameter, the null hypothesis is typically set to the value stated in the claim, while the alternative hypothesis reflects the direction of the claim. Understanding this framework is essential for conducting hypothesis tests effectively.

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Step 2: Calculate Test Statistic

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Step 2: Calculate Test Statistic Video Summary

In hypothesis testing, the primary objective is to use data from a small sample to evaluate a claim about a population. This process begins with formulating two statements: the null hypothesis (H₀) and the alternative hypothesis (H_a). For instance, if we want to test whether students at a university are younger than the previous year's average age of 23, we would set H₀: μ = 23 and H_a: μ < 23, where μ represents the population mean age.

To conduct the test, we gather a sample—in this case, data from 35 students, yielding a sample mean (x̄) of 22 years. The population standard deviation (σ) is given as 4 years. It’s crucial to differentiate between the parameters: while H₀ uses the population mean, the sample mean is derived from our collected data.

Next, we calculate a test statistic, which helps us determine how far our sample mean is from the population mean under the null hypothesis. The test statistic can be a z-score or a t-score, depending on whether the population standard deviation is known. Since we know σ, we will use the z-score formula:

z = \frac{x̄ - μ}{\frac{σ}{\sqrt{n}}}

Substituting the known values into the formula, we have:

z = \frac{22 - 23}{\frac{4}{\sqrt{35}}}

Calculating this gives us a z-score of approximately -1.479. This z-score indicates how many standard deviations the sample mean is from the population mean. A negative z-score suggests that the sample mean is below the population mean, which aligns with our alternative hypothesis.

Visualizing this on a normal distribution graph can further clarify the significance of our findings. The peak of the graph represents the population mean (23), and the calculated z-score of -1.479 shows where our sample mean (22) lies in relation to this population mean. Understanding this relationship is essential for interpreting the results of our hypothesis test.

In summary, the test statistic, whether a z-score or t-score, quantifies the difference between the sample data and the claimed population parameter, providing a basis for making informed conclusions about the hypothesis being tested.

Problem

A survey claimed that $30 percent sign$ of adults prefer electric cars over traditional cars. A car manufacturer believes the true proportion is higher than $30 percent sign$ . To test this, they survey a random sample of $50$ adults and find that $19$ say they prefer electric cars. Determine which test statistic to use & calculate it.

$z equals 1.23$

$z = - 1.23$

$z equals 4.38$

$z = - 4.38$

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Step 2: Calculate Test Statistic Example 2

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Step 2: Calculate Test Statistic Example 2 Video Summary

In this scenario, a school claims that the average score on a math final is 75. A teacher suspects that the actual average is lower, prompting the need for hypothesis testing. The null hypothesis (H₀) states that the population mean (μ) is equal to 75, while the alternative hypothesis (H_a) posits that μ is less than 75.

To test this hypothesis, a random sample of 15 students is selected. Since the sample size is less than 30 and the population standard deviation (σ) is unknown, the t-distribution is appropriate for this analysis. The test statistic is calculated using the formula:

t = \frac{\bar{x} - \mu}{s / \sqrt{n}}

Here, $\bar{x}$ represents the sample mean, μ is the hypothesized population mean (75), s is the sample standard deviation, and n is the sample size (15).

To proceed, the sample mean ($\bar{x}$) and sample standard deviation (s) must be calculated from the provided scores. After performing the calculations, the sample mean is found to be 72.4, and the sample standard deviation is approximately 3.89.

Substituting these values into the t statistic formula yields:

t = \frac{72.4 - 75}{3.89 / \sqrt{15}} ≈ -2.59

This t-score indicates how far the sample mean is from the hypothesized population mean in terms of standard deviations. A t-score of -2.59 suggests that the sample mean is significantly lower than the claimed average of 75, providing evidence against the null hypothesis.

In conclusion, the calculated t-score serves as a test statistic that helps determine whether the teacher's suspicion about the average score being lower than 75 is statistically supported.

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Step 3: Get P-Value

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Step 3: Get P-Value Video Summary

In hypothesis testing, understanding how to evaluate the significance of your sample data is crucial. After formulating your null hypothesis (H₀), which often states that there is no effect or no difference (e.g., the mean age of students is 23 years), the next step involves calculating a test statistic, such as a z-score or t-score. These scores indicate how far your sample mean deviates from the population mean under the null hypothesis.

However, simply calculating the test statistic does not determine whether to reject the null hypothesis. To make this decision, you need to calculate the p-value, which represents the probability of observing your sample data, or something more extreme, assuming the null hypothesis is true. A p-value is essentially an area under the normal distribution curve, corresponding to the test statistic.

For example, if you hypothesize that the mean age of students has decreased (H_a: μ < 23), and you calculate a z-score of -2.0, you would find the p-value by determining the area to the left of this z-score. Using a z-table or calculator, you would find that the p-value is approximately 0.0228. This low p-value indicates that the observed sample mean is quite unusual under the null hypothesis, suggesting that you may reject H₀.

In contrast, if you hypothesize that the mean age has increased (H_a: μ > 23) and obtain a z-score of 0.8, the p-value would be the area to the right of this z-score, which is approximately 0.2119. This higher p-value suggests that the sample mean is not significantly different from the null hypothesis, leading to a failure to reject H₀.

When testing for any difference in mean age (H_a: μ ≠ 23), you would conduct a two-tailed test. If you calculate a z-score of -1, the p-value would be the combined area in both tails of the distribution. This is calculated as 2 times the area to the left of -1, resulting in a p-value of approximately 0.3173. This indicates that the sample mean is not significantly different from the null hypothesis, and you would not reject H₀.

In summary, p-values are critical in hypothesis testing as they provide a measure of how unusual your sample data is under the assumption that the null hypothesis is true. A low p-value (typically less than 0.05) suggests that the sample data is unlikely under the null hypothesis, leading to its rejection, while a high p-value indicates insufficient evidence to reject the null hypothesis.

Problem

Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed.
$H sub 0 colon$ $p equals 0.4$
$H_{a} :$ $p is not equal to 0.4$

Left-tailed

Right-tailed

Two-tailed

Problem

In a certain hypothesis test, $H sub 0 colon p equals 0.4$ , $H_{a} : p$ < $0.4$ . You collect a sample and calculate a test statistic $z = - 1.32$ . Find the $P$ -value.

$P\left(z\right.$ < $- 1.32) = 0.8132$

$P\left(z\right.$ < $- 1.32) = 0.9066$

$P\left(z\right.$ < $- 1.32) = 0.0934$

$P (z$ < $- 1.32) = 0.1868$

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Step 3: Get P-Value Example 3

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Step 3: Get P-Value Example 3 Video Summary

In this example, we explore the relationship between p-values and z-scores, focusing on how to identify the correct graph that represents a given p-value of 0.2420. A p-value is a statistical measure that helps determine the significance of results in hypothesis testing, indicating the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true.

We start with a z-score of -0.82, which corresponds to the area to the left of this z-score on the standard normal distribution. The p-value in this context is represented mathematically as $ P(Z < -0.82) $. By consulting a z-table or using a graphing calculator, we find that the area associated with this z-score is approximately 0.2061. Since this value does not match our target p-value of 0.2420, we move on to the next option.

Next, we consider a right-tailed test with a z-score of 1.54. The area to the right of this z-score is represented as $ P(Z > 1.54) $. Again, using a z-table or calculator, we find that this area is approximately 0.0618, which is significantly smaller than 0.2420, indicating that this graph does not represent the correct p-value either.

Finally, we examine a z-score of -1.17 in the context of a two-tailed test. In a two-tailed test, we consider both tails of the distribution, which requires us to double the probability of the z-score being less than -1.17. This is expressed as $ 2 \times P(Z < -1.17) $. Looking up this z-score, we find that the area is approximately 0.1210. When we multiply this by 2, we obtain $ 2 \times 0.1210 = 0.2420 $, which matches our target p-value.

Thus, the correct graph representing a p-value of 0.2420 corresponds to the z-score of -1.17 in a two-tailed test, illustrating the combined area of both tails of the distribution. This example highlights the importance of understanding the relationship between z-scores, areas under the curve, and p-values in statistical analysis.

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Step 4: State Conclusion

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Step 4: State Conclusion Video Summary

In hypothesis testing, the final step involves drawing a conclusion based on the p-value obtained from the test statistics, such as z-scores or t-scores. The p-value indicates how unusual the sample data is under the assumption that the null hypothesis is true. To determine whether to reject or fail to reject the null hypothesis, we compare the p-value to a predetermined significance level, denoted as alpha (α). Common values for alpha are 0.10, 0.05, and 0.01, with 0.05 being the most frequently used.

For example, consider a researcher investigating whether the mean age of students at a university has changed from the previous year, where the mean age was 23 years. The null hypothesis (H₀) states that the mean age (μ) is equal to 23. After collecting a sample, the researcher calculates a z-score of -1.43 and a p-value of 0.0764. The next step is to compare this p-value to the significance levels provided in the problem.

In the first scenario, with α = 0.10, the p-value (0.0764) is less than α. This indicates that the sample is sufficiently unusual, leading to the conclusion that we reject the null hypothesis. This suggests there is enough evidence to support the alternative hypothesis, which posits that the mean age of students has decreased.

In the second scenario, with α = 0.05, the p-value (0.0764) is greater than α. Here, the sample does not meet the threshold for unusualness, resulting in a failure to reject the null hypothesis. This implies that there is not enough evidence to conclude that the mean age of students has changed.

To summarize the conclusion of a hypothesis test, one must follow these steps: first, draw a graph to visualize the relationship between the p-value and alpha; second, state whether to reject or fail to reject the null hypothesis based on this comparison; and finally, restate the alternative hypothesis in plain language, indicating whether there is enough evidence to support it. If the p-value is less than alpha, it suggests sufficient evidence for the alternative hypothesis; if it is greater than or equal to alpha, it indicates insufficient evidence.

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Step 4: State Conclusion Example 4

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Step 4: State Conclusion Example 4 Video Summary

In hypothesis testing, we often start by establishing a null hypothesis and an alternative hypothesis. In this case, the college organization claims that more than 10% of students read print newspapers. Therefore, the null hypothesis (H₀) is that the population proportion (p) is equal to 0.10, while the alternative hypothesis (H_a) posits that p is greater than 0.10.

To test this hypothesis, a sample is taken, and a p-value is calculated. In this scenario, the p-value obtained is 0.1632. The significance level (α) is set at 0.05, which serves as a threshold for determining whether to reject the null hypothesis. The p-value indicates the probability of observing a sample proportion as extreme as the one obtained, assuming the null hypothesis is true.

When comparing the p-value to the significance level, we find that 0.1632 is greater than 0.05. This comparison leads us to conclude that we fail to reject the null hypothesis. In practical terms, this means that there is not enough statistical evidence to support the claim that more than 10% of students read print newspapers.

In summary, the process of hypothesis testing involves calculating a p-value from sample data, comparing it to a predetermined significance level, and making a decision about the null hypothesis. In this case, the evidence was insufficient to support the alternative hypothesis, indicating that the claim made by the college organization lacks statistical backing.

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What are the steps involved in hypothesis testing?

Hypothesis testing involves several key steps. First, you formulate the null hypothesis (H₀) and the alternative hypothesis (H_a). The null hypothesis represents a claim about a population parameter, while the alternative hypothesis challenges this claim. Next, you collect sample data and calculate a test statistic, such as a z-score or t-score, which measures how far the sample data deviates from the null hypothesis. Then, you determine the p-value, which is the probability of observing the sample data if the null hypothesis is true. Finally, you compare the p-value to a predetermined significance level (α) to decide whether to reject or fail to reject the null hypothesis. This process helps determine if there is enough evidence to support the alternative hypothesis.

How do you write null and alternative hypotheses?

Writing null and alternative hypotheses involves identifying the parameter of interest and the claim being tested. The null hypothesis (H₀) is a statement of no effect or no difference, often involving an equality (e.g., μ = 23). It represents the status quo or default assumption. The alternative hypothesis (H_a) is a statement that contradicts the null hypothesis, suggesting a new effect or difference. It can involve inequalities such as less than (<), greater than (>), or not equal to (≠). For example, if testing whether students are younger than a claimed average age, H₀ might be μ = 23, while H_a could be μ < 23. The choice of symbols depends on the research question and the direction of the test.

What is a p-value in hypothesis testing?

A p-value in hypothesis testing is a probability that measures the strength of evidence against the null hypothesis (H₀). It represents the likelihood of obtaining the observed sample data, or more extreme, assuming that the null hypothesis is true. A low p-value indicates that the observed data is unusual under the null hypothesis, suggesting that the null hypothesis may not be true. Typically, a p-value is compared to a significance level (α), such as 0.05. If the p-value is less than α, the null hypothesis is rejected, indicating that there is significant evidence to support the alternative hypothesis (H_a). Conversely, if the p-value is greater than α, there is not enough evidence to reject the null hypothesis.

How do you determine which test statistic to use in hypothesis testing?

Determining which test statistic to use in hypothesis testing depends on the type of data and the parameter being tested. Common test statistics include z-scores and t-scores. A z-score is used when the population standard deviation (σ) is known and the sample size is large (typically n > 30). The formula for the z-score is: $\frac{x_{¯} - μ}{\frac{σ}{\sqrt{n}}}$ . A t-score is used when the population standard deviation is unknown and the sample size is small. The formula for the t-score is similar to the z-score but uses the sample standard deviation (s) instead of σ. The choice between z and t depends on the availability of σ and the sample size.

What is the significance level in hypothesis testing?

The significance level, denoted as α, is a threshold used in hypothesis testing to determine whether to reject the null hypothesis (H₀). It represents the probability of making a Type I error, which is rejecting the null hypothesis when it is actually true. Common significance levels are 0.10, 0.05, and 0.01, with 0.05 being the most widely used. The choice of α depends on the context of the study and the acceptable risk of error. If the p-value obtained from the test statistic is less than α, the null hypothesis is rejected, indicating that the sample data provides sufficient evidence to support the alternative hypothesis (H_a). If the p-value is greater than α, the null hypothesis is not rejected.

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Steps in Hypothesis Testing: Videos & Practice Problems

Step 1: Write Hypotheses

Step 1: Write Hypotheses Video Summary

A popular theme park claims that their weekly attendance is around 100,000100,000. You believe that the weekly attendance is different than this claimed value, so you gather sample data. Write the null and alternative hypotheses.

A candy manufacturer seeking to minimize the variation in weights of their candies claims to produce candies with a standard deviation less than 0.3000.300 g. Write the null and alternative hypotheses.

Step 1: Write Hypotheses Example 1

Step 1: Write Hypotheses Example 1 Video Summary

Step 2: Calculate Test Statistic

Step 2: Calculate Test Statistic Video Summary

Step 2: Calculate Test Statistic Example 2

Step 2: Calculate Test Statistic Example 2 Video Summary

Step 3: Get P-Value

Step 3: Get P-Value Video Summary

Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed.H0:H_0: p=0.4p=0.4Ha:H_{a}: p≠0.4p\ne0.4

In a certain hypothesis test, H0:p=0.4H_0:p=0.4, Ha:pH_{a}:p < 0.40.4. You collect a sample and calculate a test statistic z=−1.32z=-1.32. Find the PP-value.

Step 3: Get P-Value Example 3

Step 3: Get P-Value Example 3 Video Summary

Step 4: State Conclusion

Step 4: State Conclusion Video Summary

Step 4: State Conclusion Example 4

Step 4: State Conclusion Example 4 Video Summary

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Here’s what students ask on this topic:

What are the steps involved in hypothesis testing?

How do you write null and alternative hypotheses?

What is a p-value in hypothesis testing?

How do you determine which test statistic to use in hypothesis testing?

What is the significance level in hypothesis testing?

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A popular theme park claims that their weekly attendance is around $100 comma 000$ . You believe that the weekly attendance is different than this claimed value, so you gather sample data. Write the null and alternative hypotheses.

A candy manufacturer seeking to minimize the variation in weights of their candies claims to produce candies with a standard deviation less than $0.300$ g. Write the null and alternative hypotheses.

Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed.
$H sub 0 colon$ $p equals 0.4$
$H_{a} :$ $p is not equal to 0.4$

In a certain hypothesis test, $H sub 0 colon p equals 0.4$ , $H_{a} : p$ < $0.4$ . You collect a sample and calculate a test statistic $z = - 1.32$ . Find the $P$ -value.