Performing Hypothesis Tests: Means: Videos & Practice Problems

In hypothesis testing, we formulate null and alternative hypotheses to assess claims about population parameters. When the population standard deviation (σ) is known, we use the z-test; when unknown, we apply the t-test, substituting the sample standard deviation (s) and calculating the t-score. The p-value, representing the probability of observing our sample statistic under the null hypothesis, is compared to the significance level (α) to determine whether to reject the null hypothesis. Key conditions include having a random sample and a normal distribution or a sample size greater than 30.

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 hours. If an agency suspects that the actual lifespan is lower, the alternative hypothesis would be H_a: μ < 25,000 hours.

To conduct the test, we need to calculate the test statistic, which, when the population standard deviation (σ) is known, is done using the z-score formula:

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

In this scenario, the sample mean (x̄) is 24,600 hours, the population standard deviation (σ) is 1,200 hours, and the sample size (n) is 36. Plugging these values into the formula gives:

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

Next, we determine the 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 left-tailed test (since we are testing if the mean is less than 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 indicates 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 example, both conditions were satisfied, as the sample was random and the distribution was normal.

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 start by establishing our null hypothesis (H₀) and alternative hypothesis (H_a). For this example, we are testing the claim that the population mean (μ) equals 50. The null hypothesis is 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 formula for the z-score:

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

In our first scenario, the sample mean (x̄) is 47. Plugging in the values, we have:

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

Next, we determine the p-value, which represents the probability of observing a value as extreme as the test statistic under the null hypothesis. For a z-score of -3.0, 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 support the claim that the population mean is less than 50.

In the second scenario, 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 calculate the z-score as follows:

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

For a z-score of 1.0, the p-value is approximately 0.159. Since this p-value is greater than our significance level (α = 0.1), we fail to reject the null hypothesis. This suggests that there is not enough evidence to support the claim 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 draw conclusions about the population mean based on sample data.

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)

Standard Deviation (σ) Known Example 2

In this scenario, a company has historically priced one of its best-selling products at \$48. A manager suspects that the average price across different retail outlets has changed, so they take a random sample of 32 stores, which reveals an average selling price of \$46.90. The known population standard deviation is \$3.50. Given this information, the manager aims to test the hypothesis that the average price is different from the historical price of \$48.

The first step in hypothesis testing involves establishing the null hypothesis (H₀) and the alternative hypothesis (H_a). The null hypothesis assumes that the average price has not changed, thus H₀: μ = 48. The alternative hypothesis, which reflects the manager's suspicion, is that the average price is different from \$48, represented as H_a: μ ≠ 48.

Since the population standard deviation is known, the test statistic will be calculated using the z-score formula:

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

Here, \(\bar{x}\) is the sample mean (\$46.90), μ is the population mean (\$48), σ is the population standard deviation (\$3.50), and n is the sample size (32). Plugging in the values, we calculate:

z = \(\frac{46.90 - 48}{3.50 / \sqrt{32}} \approx -1.78\)

Since this is a two-tailed test (indicated by the not equal to sign in the alternative hypothesis), we need to find the p-value associated with the z-score of -1.78. The p-value represents the probability of observing a sample mean as extreme as the one obtained, under the null hypothesis. For a two-tailed test, the p-value is calculated as:

p-value = 2 × P(Z < -1.78)

Calculating this gives a p-value of approximately 0.075. This value indicates the probability of obtaining a sample mean that is either significantly lower or higher than the historical average of \$48.

The final step is to compare the p-value to the significance level (α = 0.05). Since the p-value (0.075) is greater than α (0.05), we fail to reject the null hypothesis. This outcome suggests that there is not enough evidence to conclude that the average price has changed from the historical price of \$48.

In summary, the analysis indicates that the sample data does not provide sufficient evidence to support the claim that the average price of the product is different from \$48.

Standard Deviation (σ) Known Example 3

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

Upon calculating, the test statistic is found to be approximately 0.932.

The next step is to determine the p-value associated with this z-score. The 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 the average salary has increased in the city.

In conclusion, the analysis suggests that the data does not provide sufficient evidence to assert that salaries have risen above the claimed average of \$51,000.

Standard Deviation (σ) Unknown

In hypothesis testing, when the population standard deviation (σ) is unknown, the process remains largely similar to when σ 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 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, H₀: μ = 12 (the company's claim), and H₁: μ < 12 (your suspicion).

Next, instead of using the z-score, which is applicable when σ is known, you will use the t-score because σ is unknown. 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 population mean under the null hypothesis, 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:

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

This calculation yields a t-score of approximately -3.16. To determine the p-value associated with this t-score, you need to know the degrees of freedom, which is calculated as 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 graphing calculator, you can find the p-value corresponding to the t-score of -3.16 with 39 degrees of freedom. This p-value is approximately 0.0015, indicating a very small probability of observing such a sample mean if the null hypothesis were true.

Finally, you compare the p-value to your significance level (α), which is set at 0.05. Since 0.0015 is less than 0.05, you reject the null hypothesis. This suggests that there is sufficient evidence to support the claim that the average battery life is indeed less than twelve hours.

In summary, when conducting hypothesis tests with an unknown population standard deviation, the key steps include formulating the hypotheses, calculating the t-score using the sample standard deviation, determining the p-value, and making a decision based on the comparison of the p-value to the significance level. Ensuring that the sample is random and that the sample size is adequate (n > 30) or that the underlying distribution is normal is crucial for the validity of the test.

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 4

In this scenario, we explore a hypothesis testing situation involving the average monthly rent for a one-bedroom apartment in a fictional city. The city government claims that the average rent is \$1,450, while a housing advocacy group believes this figure may be outdated. To investigate, they collect a random sample of 18 apartments, yielding a sample mean of \$1,525 and a sample standard deviation of \$135. Since the population standard deviation is unknown, we will conduct a hypothesis test for the mean using a significance level of α = 0.05.

The first step in hypothesis testing is to establish the null and alternative hypotheses. The null hypothesis (H₀) posits that the average monthly rent is \$1,450 (μ = \$1,450). The alternative hypothesis (H_a) suggests that the average rent has changed, which is represented as μ ≠ \$1,450. This indicates that we will perform a two-tailed test.

Next, we calculate the test statistic using the formula for the t-test, which is appropriate when the population standard deviation is unknown. The formula is given by:

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

Here, \(\bar{x}\) is the sample mean (\$1,525), μ is the population mean (\$1,450), s is the sample standard deviation (\$135), and n is the sample size (18). Plugging in the values, we find:

t = \frac{1525 - 1450}{135 / \sqrt{18}} ≈ 2.36

With the test statistic calculated, we now determine the p-value. For a two-tailed test, we need to find the probability of observing a t-value as extreme as 2.36 in both tails of the t-distribution. The degrees of freedom (df) for our test is calculated as n - 1, which gives us df = 17. Using a t-distribution table or calculator, we find the probability associated with t = 2.36, which is approximately 0.031. Since this is a two-tailed test, we multiply this probability by 2, resulting in a p-value of 0.062.

Finally, we compare the p-value to our significance level (α = 0.05). Since 0.031 < 0.05, we reject the null hypothesis. This indicates that there is sufficient evidence to conclude that the average monthly rent has changed, although we cannot determine whether it has increased or decreased. In summary, the analysis suggests that the claim of the average rent being \$1,450 may no longer hold true, warranting further investigation into current rental prices.

Here’s what students ask on this topic:

What are the steps to perform a hypothesis test for means when the population standard deviation is known?

To perform a hypothesis test for means when the population standard deviation (σ) is known, follow these steps:

1. State the hypotheses: Define the null hypothesis (H₀) and the alternative hypothesis (H_a). For example, H₀: μ = μ₀ and H_a: μ ≠ μ₀ (or < or >).
2. Choose the significance level (α): Common values are 0.05 or 0.01.
3. Calculate the test statistic: Use the z-test formula:$z=\frac{(x_{¯}-\mu )}{\frac{\sigma }{\sqrt{n}}}$, where x̄ is the sample mean, μ is the hypothesized mean, σ is the population standard deviation, and n is the sample size.
4. Find the p-value: Use the z-score to determine the p-value from a z-table or calculator.
5. Compare p-value to α: If p-value < α, reject H₀. Otherwise, fail to reject H₀.
6. State the conclusion: Summarize the results in the context of the problem.

How do you perform a hypothesis test for means when the population standard deviation is unknown?

When the population standard deviation (σ) is unknown, follow these steps for a hypothesis test:

1. State the hypotheses: Define the null hypothesis (H₀) and the alternative hypothesis (H_a), e.g., H₀: μ = μ₀ and H_a: μ ≠ μ₀ (or < or >).
2. Choose the significance level (α): Common values are 0.05 or 0.01.
3. Calculate the test statistic: Use the t-test formula:$t=\frac{(x_{¯}-\mu )}{\frac{s}{\sqrt{n}}}$, where x̄ is the sample mean, μ is the hypothesized mean, s is the sample standard deviation, and n is the sample size.
4. Determine degrees of freedom (df): df = n - 1.
5. Find the p-value: Use the t-score and df to find the p-value from a t-table or calculator.
6. Compare p-value to α: If p-value < α, reject H₀. Otherwise, fail to reject H₀.
7. State the conclusion: Summarize the results in the context of the problem.

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

The main difference between a z-test and a t-test lies in the availability of the population standard deviation (σ) and the sample size:

• z-test: Used when σ is known and the sample size is large (n ≥ 30). The test statistic is calculated using the formula:$z=\frac{(x_{¯}-\mu )}{\frac{\sigma }{\sqrt{n}}}$.
• t-test: Used when σ is unknown, and the sample standard deviation (s) is used instead. It is also preferred for small sample sizes (n < 30). The test statistic is calculated using the formula:$t=\frac{(x_{¯}-\mu )}{\frac{s}{\sqrt{n}}}$.
• Distribution: The z-test uses the standard normal distribution, while the t-test uses the t-distribution, which accounts for additional variability in smaller samples.

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

The p-value in hypothesis testing represents the probability of observing the sample data, or something more extreme, assuming the null hypothesis (H₀) is true. Here's how to interpret it:

• Small p-value (p < α): If the p-value is less than the significance level (α), reject H₀. This suggests that the sample data provides sufficient evidence to support the alternative hypothesis (H_a).
• Large p-value (p ≥ α): If the p-value is greater than or equal to α, fail to reject H₀. This indicates that the sample data does not provide enough evidence to support H_a.

For example, if α = 0.05 and the p-value = 0.02, you would reject H₀ because 0.02 < 0.05, suggesting strong evidence for H_a.

What are the assumptions for performing a hypothesis test for means?

When performing a hypothesis test for means, the following assumptions must be met:

• Random Sampling: The data must be collected using a random sampling method to ensure representativeness.
• Normality: The population distribution should be approximately normal. For large samples (n ≥ 30), the Central Limit Theorem allows the sample mean to be treated as normally distributed, even if the population is not.
• Independence: Observations in the sample must be independent of each other.
• Known or Unknown σ: If σ is known, use a z-test; if unknown, use a t-test with the sample standard deviation (s).

Violating these assumptions may lead to inaccurate results, so always verify them before conducting the test.