Performing Hypothesis Tests: Means: Videos & Practice Problems

In hypothesis testing, the null hypothesis (H₀) represents the default assumption, while the alternative hypothesis (H_a) suggests a different claim. When the population standard deviation (σ) is known, a z-test is used, while an unknown σ requires a t-test. The test statistic is calculated using the formula: $(x-\mu )/\sqrt{\left(s\right)/\sqrt{n})}$. The p-value determines the significance of results against a predefined alpha level.

Standard Deviation (σ) Known

In hypothesis testing, the process begins with formulating two statements: the null hypothesis (H₀) and the alternative hypothesis (H_a). The null hypothesis typically represents a statement of no effect or no difference, while the alternative hypothesis reflects the claim being tested. For instance, if a lighting company claims that their LED bulbs last an average of 25,000 hours, the null hypothesis would be H₀: μ = 25,000. If an agency suspects that the actual lifespan is lower, the alternative hypothesis would be H_a: μ < 25,000.

To conduct the test, we calculate the test statistic using the known population standard deviation (σ). In this case, the population standard deviation is 1,200 hours, and a random sample of 36 bulbs yields a sample mean (x̄) of 24,600 hours. The formula for the z-test statistic when the population standard deviation is known is given by:

\[ z = \frac{x̄ - μ}{\frac{σ}{\sqrt{n}}} \]

Substituting the values, we have:

\[ z = \frac{24,600 - 25,000}{\frac{1,200}{\sqrt{36}}} = \frac{-400}{200} = -2.00 \]

This z-score indicates how many standard deviations the sample mean is from the hypothesized population mean. Next, we convert this z-score into a p-value, which represents the probability of observing a sample mean as extreme as the one obtained, assuming the null hypothesis is true. For a one-tailed test where H_a: μ < 25,000, we look for the area to the left of z = -2.00. Using a z-table or calculator, we find that the p-value is approximately 0.023.

To make a decision, we compare the p-value to the significance level (α), which in this case is set at 0.1. Since 0.023 < 0.1, we reject the null hypothesis. This suggests that there is sufficient evidence to support the claim that the true mean lifespan of the bulbs is less than 25,000 hours.

When conducting hypothesis tests, it is crucial to ensure that certain conditions are met: the sample must be random, the population from which the sample is drawn should be normally distributed, or the sample size should be greater than 30 to invoke the Central Limit Theorem. In this scenario, both conditions were satisfied, as the distribution of bulb lifespans was normal and the sample size was 36.

In summary, hypothesis testing involves formulating hypotheses, calculating a test statistic, determining a p-value, and making a decision based on the comparison of the p-value to the significance level. This structured approach allows researchers to draw conclusions about population parameters based on sample data.

Standard Deviation (σ) Known Example 1

In hypothesis testing, we begin by establishing our null hypothesis (H₀) and alternative hypothesis (H_a). In this scenario, we are testing the claim that the population mean (μ) equals 50. The null hypothesis is set as H₀: μ = 50, while the alternative hypothesis for a left-tailed test is H_a: μ < 50.

Given the population standard deviation (σ) of 6, a sample size (n) of 36, and a significance level (α) of 0.1, we can calculate the test statistic using the z-score formula:

\[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]

For a sample mean (x̄) of 47, the calculation becomes:

\[ z = \frac{47 - 50}{6 / \sqrt{36}} = \frac{-3}{1} = -3.0 \]

Next, we determine the p-value, which represents the probability of observing a z-score less than -3.0. Using a z-table or calculator, we find that the p-value is approximately 0.0013. Since this p-value is less than our significance level (α = 0.1), we reject the null hypothesis, indicating that the sample provides sufficient evidence to suggest that the population mean is less than 50.

In the second part of the analysis, we conduct a right-tailed test with a sample mean of 51. The null hypothesis remains the same (H₀: μ = 50), but the alternative hypothesis changes to H_a: μ > 50. We again use the z-score formula:

\[ z = \frac{51 - 50}{6 / \sqrt{36}} = \frac{1}{1} = 1.0 \]

For this z-score, we calculate the p-value, which represents the probability of observing a z-score greater than 1.0. This yields a p-value of approximately 0.159. Since this p-value is greater than our significance level (α = 0.1), we fail to reject the null hypothesis, suggesting that there is not enough evidence to conclude that the population mean is greater than 50.

In summary, hypothesis testing involves setting up the null and alternative hypotheses, calculating the test statistic, determining the p-value, and comparing it to the significance level to make a decision regarding the null hypothesis. This process allows us to draw conclusions based on sample data in relation to population parameters.

Test the claim about the population mean $\mu$ at the given level of significance. Assume the population is normally distributed. Find the $P$-value and determine whether you should reject or fail to reject the null hypothesis.
Claim: $\mu \ne 1020$, $\alpha =0.01$, $\sigma =85$
Sample: $\bar{x}=990$, $n=40$

• A. P-val: 0.026; Reject the null hypothesis
• B. P-val: 0.026; Fail to reject the null hypothesis
• C. P-val: 0.013; Reject the null hypothesis
• D. P-val: 0.013; Fail to reject the null hypothesis

Answer: (B)

A university claims that the average SAT math score of its incoming freshmen is 600. A skeptical education researcher believes this might not be accurate. The researcher collects a random sample of 40 students and finds a sample mean SAT math score of 622. The population standard deviation is known to be 70. Using a significance level of $\alpha$ = 0.05, test the researcher’s claim.

• A. Since P-val > $\alpha$, we fail to reject null hypothesis. There is NOT enough evidence that SAT math scores do NOT have an average of 600.
• B. Since P-val < $\alpha$, we fail to reject null hypothesis. There is NOT enough evidence that SAT math scores do NOT have an average of 600.
• C. Since P-val > $\alpha$, we reject null hypothesis. There is enough evidence that SAT math scores do NOT have an average of 600.
• D. Since P-val < $\alpha$, we reject null hypothesis. There is enough evidence that SAT math scores do NOT have an average of 600.

Answer: (D)

Standard Deviation (σ) Known Example 2

In this scenario, we are examining the average annual salary of workers in a city, which is claimed to be \$51,000. A local labor expert believes this average has increased, prompting a hypothesis test. A random sample of 18 full-time workers is taken, and the population is assumed to be approximately normal with a known standard deviation of \$4,500. The significance level for the test is set at α = 0.05.

The first step in hypothesis testing involves formulating the null and alternative hypotheses. The null hypothesis (H₀) states that the average salary (μ) is equal to \$51,000, while the alternative hypothesis (H_a) posits that the average salary has increased, expressed as μ > \$51,000.

Next, we calculate the test statistic using the formula for the z-score:

z = \(\frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\)

Here, \(\bar{x}\) is the sample mean, μ is the population mean, σ is the standard deviation, and n is the sample size. The sample mean is calculated from the provided dataset, yielding a value of approximately 51,988.9.

Substituting the known values into the formula gives:

z = \(\frac{51,988.9 - 51,000}{4,500 / \sqrt{18}}\)

After performing the calculations, the resulting z-score is approximately 0.932.

The next step is to determine the p-value associated with this z-score. This p-value represents the probability of observing a value greater than the calculated z-score under the null hypothesis. Using statistical tables or calculators, we find that the p-value is approximately 0.176.

Finally, we compare the p-value to the significance level (α = 0.05). Since 0.176 is greater than 0.05, we fail to reject the null hypothesis. This outcome indicates that there is not enough evidence to support the claim that salaries have increased in the city.

In summary, the analysis concludes that the data does not provide sufficient evidence to suggest an increase in average salaries among workers in the city.

Standard Deviation (σ) Unknown

In hypothesis testing, when the population standard deviation (σ) is unknown, the process remains largely similar to when it is known, with a few key adjustments. This scenario is more reflective of real-world situations, as we often do not have access to the true population standard deviation.

To illustrate this, consider a tech company that claims the average battery life of its new smartphone model is twelve hours. If you suspect that the actual average is less, you would conduct a hypothesis test. You would start by formulating your null hypothesis (H₀) and alternative hypothesis (H₁). In this case, the null hypothesis would state that the average battery life (μ) is equal to twelve hours (H₀: μ = 12), while the alternative hypothesis would suggest that it is less than twelve hours (H₁: μ < 12).

When the population standard deviation is unknown, the t-distribution is used instead of the normal distribution. The t-score is calculated using the formula:

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

Here, \bar{x} is the sample mean, μ is the hypothesized population mean, s is the sample standard deviation, and n is the sample size. For example, if you take a sample of 40 smartphones and find a sample mean battery life of 11.4 hours with a sample standard deviation of 1.2 hours, you would substitute these values into the formula:

t = \frac{11.4 - 12}{1.2 / \sqrt{40}} = -3.16

Next, to determine the p-value associated with this t-score, you need to calculate the degrees of freedom, which is given by n - 1. In this case, with a sample size of 40, the degrees of freedom would be 39. Using the t-distribution table or a calculator, you can find the p-value corresponding to the t-score of -3.16, which is approximately 0.0015.

To make a decision regarding the null hypothesis, compare the p-value to the significance level (α), which is typically set at 0.05. Since 0.0015 is less than 0.05, you reject the null hypothesis, concluding that there is sufficient evidence to support the claim that the average battery life is less than twelve hours.

In summary, when conducting hypothesis tests with an unknown population standard deviation, it is crucial to use the t-distribution and calculate the t-score and p-value accordingly. Ensure that the sample is random and that either the sample size is greater than 30 or the underlying distribution is normal to validate the test results.

Test the claim about the population mean $\mu$ at the given level of significance. Assume the population is normally distributed. Find the $P$-value and determine whether you should reject or fail to reject the null hypothesis.
Claim: $\mu >52$, $\alpha =0.10$
Sample: $\bar{x}=53.1$, $s=4.7$, $n=20$

• A. P-val = 0.154; Reject the null hypothesis.
• B. P-val = 0.154; Fail to reject the null hypothesis.
• C. P-val = 0.308; Reject the null hypothesis.
• D. P-val = 0.308; Fail to reject the null hypothesis.

Answer: (B)

Standard Deviation (σ) Unknown Example 3

In hypothesis testing, researchers often start with a claim that needs to be tested. In this case, a researcher claims that the average height of adult males in a city is 70 inches. To evaluate this claim, a random sample of 12 adult males is taken, yielding a sample mean height of 67.8 inches and a sample standard deviation of 2.1 inches. The significance level (alpha) is set at 0.05, a common threshold for determining statistical significance.

The first step in hypothesis testing involves formulating the null hypothesis (H₀) and the alternative hypothesis (H_a). Here, the null hypothesis states that the average height (μ) is equal to 70 inches (H₀: μ = 70). The alternative hypothesis, which challenges the null, is that the average height is not equal to 70 inches (H_a: μ ≠ 70). This indicates a two-tailed test, as we are interested in deviations in both directions from the claimed mean.

Next, the test statistic is calculated using the t-test formula, which is appropriate here since the population standard deviation is unknown. The formula for the t-test statistic is given by:

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

Substituting the values, we have:

t = \frac{67.8 - 70}{2.1 / \sqrt{12}} ≈ -3.63

This t-score indicates how many standard deviations the sample mean is from the hypothesized population mean. To determine the p-value, which helps in making a decision about the null hypothesis, we need to find the probability associated with this t-score. Since this is a two-tailed test, we will multiply the one-tailed p-value by two.

The degrees of freedom (df) for the t-test is calculated as the sample size minus one, which in this case is 11 (df = 12 - 1). Using a t-distribution table or calculator, we find the probability for t < -3.63. This results in a very small p-value, approximately 0.002. Multiplying by two for the two-tailed test gives a final p-value of about 0.004.

Finally, we compare the p-value to the significance level (alpha). Since the p-value (0.004) is less than alpha (0.05), we reject the null hypothesis. This suggests that there is sufficient evidence to conclude that the average height of adult males in the city is not equal to 70 inches. In summary, the results indicate that the claim of an average height of 70 inches is unlikely to be true based on the sample data.

Standard Deviation (σ) Unknown Example 4

In this example, we explore a hypothesis test concerning the average resting heart rate of yoga instructors compared to the general adult population, which has an average resting heart rate of 72 beats per minute. The researcher claims that yoga instructors have a lower average resting heart rate. To test this claim, we will conduct a hypothesis test using a sample of 10 randomly selected individuals.

We begin by establishing our hypotheses. The null hypothesis (H₀) posits that the average resting heart rate (μ) is equal to 72, while the alternative hypothesis (H_a) suggests that the average resting heart rate is less than 72, expressed as μ < 72.

Since the population standard deviation is unknown, we will utilize the t-test statistic for our calculations. The formula for the t-test statistic is given by:

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

In this formula, \(\bar{x}\) represents the sample mean, μ is the population mean, s is the sample standard deviation, and n is the sample size. In this case, we need to calculate the sample mean and standard deviation from the provided data. After performing the necessary calculations, we find that the sample mean (\(\bar{x}\)) is 68.4 and the sample standard deviation (s) is 5.21.

Substituting these values into the t-test formula, we have:

t = \(\frac{68.4 - 72}{5.21 / \sqrt{10}}\)

Calculating this gives us a t-score of approximately -2.18. Next, we need to determine the p-value associated with this t-score. With 9 degrees of freedom (n - 1 = 10 - 1), we can look up the p-value in a t-distribution table or use a calculator. The resulting p-value is 0.028.

Finally, we compare the p-value to our significance level (α = 0.05). Since 0.028 is less than 0.05, we reject the null hypothesis. This indicates that there is sufficient evidence to support the claim that the average resting heart rate of yoga instructors is indeed lower than that of the general adult population.

In summary, through this hypothesis testing process, we have demonstrated how to analyze sample data to draw conclusions about population parameters, reinforcing the importance of statistical methods in research.

Here’s what students ask on this topic:

What is the difference between a z-test and a t-test in hypothesis testing?

A z-test is used when the population standard deviation (σ) is known, while a t-test is applied when σ is unknown and the sample standard deviation (s) is used instead. The z-test relies on the standard normal distribution, whereas the t-test uses the t-distribution, which accounts for additional variability in smaller samples. The formula for the z-test statistic is:

$\frac{(x-\mu )}{\frac{\sigma }{\sqrt{n}}}$

For the t-test, the formula is similar but replaces σ with s:

$\frac{(x-\mu )}{\frac{s}{\sqrt{n}}}$

The t-distribution becomes closer to the normal distribution as the sample size increases.

How do you calculate the p-value in hypothesis testing?

The p-value represents the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. For a z-test, the p-value is determined using the z-score and the standard normal distribution. For a t-test, the p-value is calculated using the t-score and the t-distribution, which requires degrees of freedom (df = n - 1). You can use statistical tables, software, or calculators to find the p-value. For example, if the z-score is -2.00, the p-value corresponds to the area to the left of -2.00 in the standard normal distribution. Similarly, for a t-score of -3.16 with df = 39, the p-value is found using the t-distribution table or software.

What are the steps to perform a hypothesis test for means?

To perform a hypothesis test for means, follow these steps:

1. State the null (H₀) and alternative (H_a) hypotheses. For example, H₀: μ = 25,000 and H_a: μ < 25,000.

2. Choose the test type (z-test or t-test) based on whether the population standard deviation (σ) is known.

3. Calculate the test statistic using the formula:

   $\frac{(x-\mu )}{\frac{\sigma }{\sqrt{n}}}$

   or

   $\frac{(x-\mu )}{\frac{s}{\sqrt{n}}}$

4. Find the p-value using the test statistic and distribution.

5. Compare the p-value to the significance level (α). If p < α, reject H₀; otherwise, fail to reject H₀.

6. State the conclusion in context.

What conditions must be met to perform a hypothesis test for means?

To perform a hypothesis test for means, the following conditions must be met:

1. The sample must be random to ensure unbiased results.

2. The population distribution should be normal, or the sample size (n) should be large enough (n > 30) to approximate normality using the Central Limit Theorem.

3. If using a t-test, the sample standard deviation (s) must be available, and degrees of freedom (df = n - 1) must be calculated.

4. The significance level (α) should be predetermined, typically 0.05 or 0.1.

These conditions ensure the validity and reliability of the hypothesis test results.

How do you interpret the results of a hypothesis test?

To interpret the results of a hypothesis test:

1. Compare the p-value to the significance level (α). If p < α, reject the null hypothesis (H₀), indicating sufficient evidence for the alternative hypothesis (H_a).

2. If p ≥ α, fail to reject H₀, meaning there is insufficient evidence to support H_a.

3. State the conclusion in context. For example, if testing whether μ < 25,000 and p < α, conclude that there is evidence to suggest the mean is less than 25,000.

Always ensure the conclusion aligns with the hypotheses and the problem's context.