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

z = − 1.23 z=-1.23 z = − 1.23

z = − 4.38 z=-4.38 z = − 4.38

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

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

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

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

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

Topic summary

In hypothesis testing, the process begins with formulating 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. After establishing these hypotheses, a test statistic, such as a z-score or t-score, is calculated to assess how unusual the sample data is under the null hypothesis. The p-value, a probability indicating the likelihood of observing the sample data if H₀ is true, is then compared to a significance level (α) to determine whether to reject or fail to reject H₀.

concept

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_n $ or $ H_0 $), represents a claim about a population parameter that we assume to be true. This parameter can be a mean ($ \mu $), a proportion ($ p $), or a standard deviation ($ \sigma $), and it is typically expressed with an equal sign. For instance, if a claim states that 30% of people like vanilla ice cream, the null hypothesis would be $ H_0: p = 0.3 $.

In contrast, the alternative hypothesis, represented as $ H_a $, is the opposing claim that we seek evidence for. It shares the same parameter as the null hypothesis but differs in the relational symbol, which can be less than (<), greater than (>), or not equal to (≠). For example, if we want to test whether the mean age of students is less than the claimed average of 23 years, the null hypothesis would be $ H_0: \mu = 23 $ and the alternative hypothesis would be $ H_a: \mu < 23 $.

Understanding the context of the problem is vital for correctly identifying these hypotheses. Keywords such as "greater than," "less than," or "not equal to" help determine the appropriate relational symbol for the alternative hypothesis. For instance, if a business journal claims that more than 20% of companies have female CEOs, the null hypothesis would be $ H_0: p = 0.2 $ and the alternative hypothesis would be $ H_a: p > 0.2 $.

These hypotheses serve as the foundation for statistical testing, allowing researchers to challenge the null hypothesis based on sample data. The null hypothesis provides a benchmark for comparison, while the alternative hypothesis indicates the direction of the test. By clearly defining these statements, researchers can effectively analyze data and draw meaningful conclusions.

Problem

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

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

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

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

$H_{a}:$ $\mu\ne100,000$

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

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

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

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

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.

$H_0:$ $\sigma=0.300$ g

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

$H_0:$ $\sigma=0.300$ g

$H_{a}:$ $\sigma\ne0.300$ g

$H_0:$ $\sigma=0.300$ g

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

$H_0:$ $\sigma>0.300$ g

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

example

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 that the population mean, denoted as $ \mu $, is less than 120. This claim serves as the basis for our alternative hypothesis.

The null hypothesis, represented as $ H_0 $, is a statement of no effect or no difference, and it typically asserts that a parameter equals a specific value. In this case, since the claim involves $ \mu < 120 $, the null hypothesis will be that $ \mu $ is equal to 120, or:

\( H_0: \mu = 120 \

Conversely, the alternative hypothesis, denoted as $ H_a $, reflects the claim we are testing. Here, it states that the population mean is indeed less than 120, which can be expressed as:

\( H_a: \mu < 120 \

In summary, when faced with a claim about a parameter, the null hypothesis is formulated as the parameter being equal to a specific value, while the alternative hypothesis represents the claim being tested. This structure is essential for conducting hypothesis tests and drawing conclusions based on statistical evidence.

Problem

Student loan debt has fluctuated over years, with signs indicating that the default rate may be increasing. Write the null and alternative hypothesis if you want to determine if the student loan default rate this year is more than $15 percent sign$ .

$H sub 0 colon p equals 0.15$

$H sub a colon p is greater than 0.15$

$H sub 0 colon p equals 0.85$

$H sub a colon p is greater than 0.85$

$H sub 0 colon p is greater than 0.15$

$H sub a colon p equals 0.15$

$H sub 0 colon p is greater than 0.85$

$H sub a colon p equals 0.85$

concept

Step 2: Calculate Test Statistic

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Step 2: Calculate Test Statistic Video Summary

In hypothesis testing, the primary objective is to evaluate a claim about a population using data from a smaller sample. This process begins with formulating two key statements: the null hypothesis (H₀) and the alternative hypothesis (H_a). For instance, if we want to determine 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, 35 students—and calculate the sample mean (x̄). If our sample yields a mean age of 22 years, we must proceed with caution. Although 22 is less than 23, it is essential to assess whether this difference is statistically significant or merely a result of sampling variability.

To quantify this, we calculate a test statistic, which can be a z-score or a t-score, depending on whether the population standard deviation (σ) is known. In this scenario, we know σ = 4 years, allowing us to 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 far the sample mean is from the population mean in terms of standard deviations. A negative z-score suggests that the sample mean is below the population mean, which is consistent with our alternative hypothesis.

Visualizing this on a normal distribution graph helps contextualize the z-score. The peak of the graph represents the population mean (23), and the calculated z-score of -1.479 shows where the sample mean (22) lies relative to this mean. Understanding the significance of this z-score is crucial for determining whether to reject the null hypothesis in favor of the alternative hypothesis.

In summary, the test statistic serves as a critical tool in hypothesis testing, allowing researchers to assess the difference between sample data and the claimed population parameter. The calculated z-score of -1.479 provides insight into the relationship between the sample mean and the hypothesized population mean, guiding the decision-making process in hypothesis testing.

Problem

A survey claimed that $30\%$ of adults prefer electric cars over traditional cars. A car manufacturer believes the true proportion is higher than $30\%$ . 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.

$z=1.23$

$z=-1.23$

$z=4.38$

$z=-4.38$

example

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 a hypothesis test. 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, a random sample of 15 students is selected, and their scores are analyzed.

Since the sample size is small (n < 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). The sample mean ($\bar{x}$) is calculated from the provided scores, yielding a value of 72.4. The sample standard deviation (s) is computed to be approximately 3.89.

Substituting these values into the formula gives:

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

Calculating this results in a t-score of approximately -2.59. This negative value indicates that the sample mean is significantly lower than the hypothesized population mean of 75. The t-score serves as a standardized measure of how far the sample mean deviates from the population mean, allowing for further statistical analysis to determine if the null hypothesis can be rejected in favor of the alternative hypothesis.

In conclusion, the calculated t-score of -2.59 suggests that there is evidence to support the teacher's claim that the average score is indeed lower than 75, warranting further investigation into the population's performance on the math final.

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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 test statistics, such as z or t scores. These scores indicate how far your sample mean deviates from the population mean under the null hypothesis.

However, simply calculating a z or t score 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 essentially quantifies how unusual your sample is within the context of the null hypothesis.

For instance, if you hypothesize that the mean age of students has decreased, your alternative hypothesis (H_a) would be that the mean age is less than 23 (H_a: μ < 23). If you calculate a z score of -2.0, this indicates that your sample mean is two standard deviations below the assumed mean of 23. To find the p-value, you would look for the area to the left of this z score on the standard normal distribution. Using a z-table or calculator, you would find that the p-value is approximately 0.0228, indicating a low probability of observing such a sample mean if the null hypothesis were true. This low p-value suggests that the sample is unusual, potentially leading to the rejection of the null hypothesis in a left-tailed test.

Conversely, if your alternative hypothesis states that the mean age has increased (H_a: μ > 23), and you obtain a z score of 0.8, the p-value would represent the area to the right of this z score. In this case, the p-value is approximately 0.2119, indicating a higher probability of observing such a sample mean, which may not warrant rejecting the null hypothesis in a right-tailed test.

In scenarios where you are testing for any difference in mean age (H_a: μ ≠ 23), you would conduct a two-tailed test. If you find a z score of -1, the p-value would be calculated as the sum of the areas in both tails of the distribution. This means you would double the area corresponding to the z score of -1, resulting in a p-value of approximately 0.3173. This reflects the total probability of observing a sample mean that is either significantly lower or higher than the null hypothesis mean.

In summary, p-values are essential for determining the significance of your results in hypothesis testing. They provide a probabilistic measure of how extreme your sample data is under the assumption that the null hypothesis is true, guiding you in making informed decisions about rejecting or failing to reject the null hypothesis.

Problem

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

Left-tailed

Right-tailed

Two-tailed

Problem

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

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

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

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

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

example

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 how to match a given p-value of 0.242 with the appropriate graph that represents the corresponding area under the normal distribution curve. It's important to clarify that this p-value is not a proportion but rather a probability value associated with z-scores.

To begin, we have a z-score of -0.82. The p-value represents the area to the left of this z-score, which can be mathematically expressed as $ P(Z < -0.82) $. By consulting a z-table or using a graphing calculator, we find that the area corresponding to a z-score of -0.82 is approximately 0.2061. This value does not match our target p-value of 0.242, prompting us to examine other options.

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 the area is approximately 0.0618, which is significantly smaller than 0.242.

Finally, we analyze a z-score of -1.17 in the context of a two-tailed test. In a two-tailed test, we consider both ends of the distribution, and the p-value is calculated as twice the area to the left of the z-score. Thus, we express this as $ P(Z < -1.17) $. Looking this up, we find that the area is approximately 0.121. Multiplying this by two gives us $ 2 \times 0.121 = 0.242 $, which matches our given p-value.

In summary, the correct graph corresponds to the z-score of -1.17, as it accurately represents the combined area of both tails in the distribution, yielding the desired p-value of 0.242.

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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 calculated p-value and the predetermined significance level, denoted as alpha (α). The p-value represents the probability of obtaining a sample statistic as extreme as the one observed, assuming the null hypothesis (H₀) is true. The significance level, typically set at values like 0.1, 0.05, or 0.01, serves as a threshold to determine whether the observed data is sufficiently unusual to reject the null hypothesis.

To illustrate this process, consider a scenario where a researcher is investigating whether the mean age of students at a university has changed from the previous year's mean of 23 years. The null hypothesis states that the mean age (μ) equals 23. After collecting data, the researcher calculates a test statistic (z = -1.43) and finds a p-value of 0.0764.

When concluding the hypothesis test, the first step is to compare the p-value to the significance level. For example, if α = 0.1, since the p-value (0.0764) is less than α, the conclusion is to reject the null hypothesis. This indicates that the sample data is sufficiently unusual, suggesting that the mean age has likely changed. Conversely, if α = 0.05, the p-value exceeds this threshold, leading to a failure to reject the null hypothesis, implying that there is not enough evidence to claim a change in mean age.

In summary, the relationship between the p-value and alpha determines the outcome of the hypothesis test. If the p-value is less than alpha, the null hypothesis is rejected, indicating enough evidence to support the alternative hypothesis. If the p-value is greater than or equal to alpha, the null hypothesis is not rejected, suggesting insufficient evidence to support the alternative hypothesis. This systematic approach ensures clarity in interpreting statistical results and making informed conclusions based on the data.

example

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 0.1, while the alternative hypothesis (H_a) posits that p is greater than 0.1.

After conducting a sample and calculating a p-value of 0.1632, we compare this value to a predetermined significance level (α), which in this scenario is set at 0.05. The p-value represents the probability of observing a sample proportion as extreme as, or more extreme than, the one obtained, assuming the null hypothesis is true. A p-value of 0.1632 indicates that there is a 16.32% chance of observing such a sample if the true proportion of students reading print newspapers is indeed 10%.

To determine whether to reject or fail to reject the null hypothesis, we compare the p-value to the significance level. Since 0.1632 is greater than 0.05, we fail to reject the null hypothesis. This means that the evidence is not strong enough to support the claim that more than 10% of students read print newspapers.

In conclusion, based on the hypothesis test, we do not have sufficient evidence to suggest that the proportion of students reading print newspapers exceeds 10%. This outcome highlights the importance of understanding the relationship between p-values and significance levels in hypothesis testing.

Problem

A transportation analyst claims that the average commute time in a major city is less than $45$ minutes. A hypothesis test is conducted, with a resulting P-value of $0.0084$ . What would be the conclusion at a significance level $alpha equals .01$

We reject $H sub 0$ because we have enough evidence that the average commute time is less than $45$ minutes

We adopt $H sub 0$ because we lack sufficient evidence that the average commute time is less than $45$ minutes

No conclusion can be made

Here’s what students ask on this topic:

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, to assess how different the sample data is from the null hypothesis. Then, you determine the p-value, which is the probability of observing the sample data under the null hypothesis. Finally, you compare the p-value to a predetermined significance level (α) to decide whether to reject or fail to reject the null hypothesis. If the p-value is less than α, you reject H₀; otherwise, you fail to reject it.

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 the statement you aim to find evidence for, and it can involve inequalities (e.g., μ < 23, μ > 23, or μ ≠ 23). The choice of inequality depends on the research question. For example, if you are testing whether a mean is less than a claimed value, H_a would be μ < claimed value. These hypotheses guide the direction of the test and the interpretation of results.

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 small p-value (typically ≤ 0.05) indicates strong evidence against H₀, leading to its rejection. Conversely, a large p-value suggests weak evidence against H₀, resulting in a failure to reject it. The p-value helps determine the statistical significance of the test results. It is crucial to compare the p-value with a predetermined significance level (α) to make a decision about the null hypothesis.

How do you calculate a test statistic in hypothesis testing?

Calculating a test statistic in hypothesis testing involves using sample data to assess the difference from the null hypothesis. For a mean, the test statistic can be a z-score or t-score, depending on whether the population standard deviation (σ) is known. The formula for a z-score is:

$\frac{(x_{-} μ)}{\frac{σ}{\sqrt{n}}}$

where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. For a t-score, replace σ with the sample standard deviation (s). The test statistic quantifies how far the sample mean is from the population mean, helping to determine the p-value and the decision about the null hypothesis.

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 rejecting H₀ when it is actually true, known as a Type I error. Common significance levels are 0.05, 0.01, and 0.10. A lower α indicates a stricter criterion for rejecting H₀. When the p-value is less than or equal to α, the null hypothesis is rejected, suggesting that the sample data provides sufficient evidence against H₀. The choice of α depends on the context of the study and the acceptable risk of making a Type I error.

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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 100,000100,000100,000. 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 0.3000.3000.300 g. Write the null and alternative hypotheses.

Step 1: Write Hypotheses Example 1

Step 1: Write Hypotheses Example 1 Video Summary

Student loan debt has fluctuated over years, with signs indicating that the default rate may be increasing. Write the null and alternative hypothesis if you want to determine if the student loan default rate this year is more than 15%15\%.

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:H0: p=0.4p=0.4p=0.4Ha:H_{a}:Ha: p≠0.4p\ne0.4p=0.4

In a certain hypothesis test, H0:p=0.4H_0:p=0.4H0:p=0.4, Ha:pH_{a}:pHa:p < 0.40.40.4. You collect a sample and calculate a test statistic z=−1.32z=-1.32z=−1.32. Find the PPP-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

A transportation analyst claims that the average commute time in a major city is less than 4545 minutes. A hypothesis test is conducted, with a resulting P-value of 0.00840.0084. What would be the conclusion at a significance level α=.01\alpha=.01

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 calculate a test statistic in hypothesis testing?

What is the significance level in hypothesis testing?

Your Statistics for Business tutors

A popular theme park claims that their weekly attendance is around $100,000$ . 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.

Student loan debt has fluctuated over years, with signs indicating that the default rate may be increasing. Write the null and alternative hypothesis if you want to determine if the student loan default rate this year is more than $15 percent sign$ .

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

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

A transportation analyst claims that the average commute time in a major city is less than $45$ minutes. A hypothesis test is conducted, with a resulting P-value of $0.0084$ . What would be the conclusion at a significance level $alpha equals .01$