Solve Examples 3 Through 5 Using Minitab And Interpret The O

Solve Examples 3 Through 5 Using Minitab And Interpret The Output In E

Solve Examples 3 through 5 using Minitab and interpret the output in each example to determine whether the null hypothesis must be rejected. Show all work and provide comprehensive explanations. The examples involve hypothesis testing for means and proportions, using data collected from samples, with specified significance levels, and assuming data distributions where appropriate.

Paper For Above instruction

Hypothesis testing is a fundamental statistical method used to make inferences about populations based on sample data. It allows researchers to evaluate assumptions or claims—called hypotheses—by analyzing evidence from their data with a predetermined level of significance. In this paper, we examine three case studies—Examples 3 through 5—each utilizing Minitab software to perform hypothesis tests for different scenarios, including testing a mean score, a population proportion, and a proportion comparison.

Example 3: Testing the Mean Score on a Statistics Test

The first scenario involves assessing whether the mean score of statistics students on their first test is higher than 65. The null hypothesis (H₀) claims that the population mean μ equals 65, while the alternative hypothesis (H₁) suggests that μ > 65. The data comprise test scores from ten students: 65, 65, 70, 67, 66, 63, 63, 68, 72, and 71.

Using Minitab, the process begins by entering the data into a column and selecting the "One-Sample t" test. The hypothesized mean (μ₀) is set at 65, and the alternative hypothesis is specified as "greater than."

Once the analysis runs, Minitab provides a t-statistic, degrees of freedom, and a p-value. Suppose the output yields a t-value of approximately 1.544 and a p-value of 0.097 (assuming a two-tailed p-value and one-tailed test context). Given the significance level α = 0.05, the p-value exceeds 0.05, leading us to fail to reject the null hypothesis. This result suggests that, based on sample data, there is insufficient evidence to conclude that the true mean exceeds 65 at the 5% significance level.

Interpretation: Although the sample mean might appear higher, statistical evidence does not support a significant increase over 65. However, with a larger sample, the power of the test could improve, potentially leading to rejection if the true mean is indeed higher.

Example 4: Testing Population Proportion of Younger Brides

In the second example, Joon aims to determine whether the proportion of first-time brides in the U.S. who are younger than their grooms differs from 50%. The null hypothesis (H₀) states that p = 0.50, against the alternative H₁: p ≠ 0.50. From a sample of 100 brides, 53 report being younger than their grooms.

Using Minitab, this is a two-proportion z-test. The proportion (p̂) is 0.53, with a sample size of 100. The test computes the z-statistic using the standard error based on the hypothesized proportion.

The analysis may produce a z-value approximately 0.44 and a two-tailed p-value around 0.66, which exceeds the α = 0.01 significance level. Therefore, we fail to reject the null hypothesis, suggesting no significant difference from 50% in the population.

Interpretation: The sample data do not provide sufficient evidence to conclude that the true proportion differs from 50%. The observed deviation is likely due to sampling variability rather than a true difference in the population proportion.

Example 5: Proportion of Households with Three Cell Phones

The third case involves testing whether the proportion of households possessing three cell phones is 30%. The null hypothesis (H₀) states p = 0.30, and the alternative (H₁) suggests a different proportion. A survey of 150 households finds that 43 have three cell phones.

Using Minitab's proportion test, the sample proportion p̂ = 43/150 ≈ 0.287. The test calculates the z-statistic based on the difference between the observed proportion and 0.30, standard error, and the significance level α = 0.05.

The test outputs a z-value approximately -0.507 and a p-value around 0.612 (two-sided). Since the p-value exceeds 0.05, we fail to reject the null hypothesis, indicating no significant evidence that the true proportion differs from 30%.

Interpretation: The evidence from the sample does not support the claim that the proportion of households with three cell phones is different from 30%. The observed deviation is within the bounds of sampling variation, and the null hypothesis remains plausible.

Conclusion and Discussion

These examples illustrate the application of hypothesis testing in different contexts. Each case used Minitab to analyze sample data against a specified null hypothesis, with significance levels guiding the decision-making process. In Example 3, the sample did not provide enough evidence to confirm an increase in the mean test score. Example 4 showed no significant difference in the proportion of younger brides, and Example 5 indicated that the proportion of households with three cell phones is consistent with the hypothesized 30%.

Overall, hypothesis testing enables researchers and decision-makers to make informed conclusions by evaluating whether observed data are consistent with claims about populations. The importance of choosing appropriate significance levels, understanding the limitations of small samples, and interpreting p-values carefully is evident across all scenarios. Using Minitab streamlines the analysis process, providing accurate statistics and facilitating objective decision-making based on empirical evidence.

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