Question

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A recent random survey of households found that 14 out of 50 householders had a cat and 21 out of 60 householders had a dog. At \(\alpha=0.05,\) test the claim that fewer household owners have cats than household owners who have dogs as pets.

Short answer

There is insufficient evidence to support the claim that fewer households have cats than dogs.

Step-by-step solution

State the Hypotheses

Set up the null hypothesis and alternative hypothesis. Let \( p_1 \) be the proportion of households with cats and \( p_2 \) be the proportion of households with dogs.\[ H_0: p_1 \geq p_2 \] (Null Hypothesis)\[ H_1: p_1 < p_2 \] (Alternative Hypothesis, this is the claim)

Calculate Sample Proportions

Calculate the sample proportions of households with cats (\( \hat{p}_1 \)) and dogs (\( \hat{p}_2 \)). \[ \hat{p}_1 = \frac{14}{50} = 0.28 \]\[ \hat{p}_2 = \frac{21}{60} = 0.35 \]

Find the Critical Value

Since it's a left-tailed test, we will find the critical value for \( \alpha = 0.05 \) using the z-distribution. The critical value is approximately \( z = -1.645 \).

Compute the Test Value

Use the formula for the test statistic for two proportions:\[ z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \]First, find the pooled proportion:\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{14 + 21}{50 + 60} = \frac{35}{110} = 0.318 \]Plug these into the test value formula:\[ z = \frac{0.28 - 0.35}{\sqrt{0.318 \times 0.682 \times \left(\frac{1}{50} + \frac{1}{60}\right)}} \]\[ z \approx \frac{-0.07}{0.0948} \approx -0.738 \]

Make the Decision

Compare the test value to the critical value:Since \( z \approx -0.738 \) is greater than \( -1.645 \), we fail to reject the null hypothesis.

Summarize the Results

At \( \alpha = 0.05 \), there is not enough evidence to reject the null hypothesis. Therefore, we do not support the claim that fewer household owners have cats than owners who have dogs.

Key concepts

Null Hypothesis

The null hypothesis is a cornerstone of hypothesis testing. It serves as the default or starting assumption that there is no effect or no difference in the context of a specific claim. In the exercise, the null hypothesis (\( H_0 \)) is defined as \( p_1 \geq p_2 \). This means we assume that the proportion of households with cats (\( p_1 \)) is greater than or equal to the proportion of households with dogs (\( p_2 \)). By stating this hypothesis, we are essentially positing that any difference observed is due to random chance rather than a real difference in the populations.

• The null hypothesis is always formulated to be tested and potentially rejected.
• It provides a condition that is tested against the data collected.
• If evidence suggests deviation from the null hypothesis, it may be suitable to reject it.

Understanding the null hypothesis is crucial as it sets the baseline for our testing and influences all subsequent steps in hypothesis testing.

Alternative Hypothesis

The alternative hypothesis contrasts the null hypothesis, representing the claim that needs to be tested. In this scenario, the alternative hypothesis (\( H_1 \)) is stated as \( p_1 < p_2 \). This suggests that the proportion of households with cats is less than those with dogs, which directly aligns with the main claim of the exercise.

• The alternative hypothesis is what researchers typically want to prove or support.
• It reflects a specific proposed change or effect we are testing for.
• In our case, it's the hypothesis that fewer households have cats than dogs.

The process of hypothesis testing aims to determine if there is enough evidence to support the alternative hypothesis over the null hypothesis. Understanding the alternative hypothesis is key, as it plays a crucial role in decision-making based on the test performed.

Critical Value

The critical value is a point on the test statistic distribution that defines the threshold for statistical significance. For a left-tailed test like this, the critical value is the boundary below which we would reject the null hypothesis in favor of the alternative. In the exercise, with a significance level (\( \alpha \)) of 0.05, the critical z-value is approximately \( z = -1.645 \). This value acts as a benchmark for comparing the computed test statistic.

• Critical values depend on the chosen significance level.
• They help determine whether the observed data falls in the tail, supporting a potential rejection of the null hypothesis.
• The critical value separates the region where we would reject the null hypothesis from the region where we would not.

Comprehending the critical value allows us to understand the decision-making process in hypothesis testing by establishing a standard for evaluation.

Test Statistic

The test statistic is a standardized value used to determine whether to retain or reject the null hypothesis. It represents how far the observed data deviates from the expectation under the null hypothesis, scaled relative to the expected variability. In this exercise, we calculate the z-test statistic using the data for the two sample proportions; \( z \approx -0.738 \).

• The test statistic is derived from data collected and varies based on the test being performed.
• It's compared to the critical value to help decide on the hypothesis.
• If the test statistic falls beyond the critical value, it suggests a significant result supporting the alternative hypothesis.
• In this case, since \( z \approx -0.738 \) does not lie in the critical region, we do not reject the null hypothesis.

Understanding the test statistic gives us insights into the magnitude and significance of the observed difference in relation to the expected if the null hypothesis were true.