The Concepts Professor Nord Stated That The Mean Score On

The Conceptsprofessor Nord Stated That The Mean Score On

The assignment involves performing hypothesis testing related to population means based on sample data. It requires identifying hypotheses, choosing the correct test type, calculating test statistics and p-values, and drawing conclusions based on significance levels. The task includes analyzing different scenarios such as mean exam scores, credit balances, and home prices with varying sample sizes, sample means, standard deviations, and significance levels.

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Hypothesis testing is a central process in statistics that allows researchers to make inferences about a population parameter based on sample data. This process involves several steps: formulating hypotheses, selecting an appropriate test, calculating the test statistic and p-value, and then making a decision to reject or fail to reject the null hypothesis. In this context, we analyze three scenarios to illustrate the application of hypothesis testing principles.

Scenario 1: Exam scores and population mean

Professor Nord has stated that the average score on the final exam over several years is 79%. Colby’s class, with a mean score of 77%, seeks to determine if this difference is statistically significant. The significance level provided is α = 0.05. From this, the population mean is known (μ = 79%), and the sample mean (x̄ = 77%) is given. The question is whether this difference indicates a real variation or could be attributed to random sampling variability.

The test conducted here is two-tailed because the alternative hypothesis tests whether the sample mean is either significantly higher or lower than the population mean. This is appropriate because the question aims to determine if Colby’s class differs in either direction, not just one.

The null hypothesis (H0) assumes no difference between sample and population means: H0: μ = 79%. The alternative hypothesis (Ha) states that there is a difference: Ha: μ ≠ 79%. To test this, a z-test is typically used if the population standard deviation is known, otherwise a t-test is appropriate when standard deviation is unknown or the sample size is small.

If the calculated p-value from the test statistic is 0.09, which is less than the significance level α = 0.05, Colby should reject the null hypothesis, indicating that there is a statistically significant difference between his class’s mean score and the population mean. Conversely, if the p-value were greater than 0.05, he would fail to reject the null hypothesis, suggesting that the observed difference could be due to chance.

Colby’s concluding statement could be: “Based on the data, there is enough evidence at the 5% significance level to conclude that the mean score of my class is different from the overall population mean of 79%.”

Scenario 2: Credit account balances

A sample of 15 credit account balances shows a mean balance of $4,300 with a sample standard deviation of $400. The marketing manager claims that the population mean balance should be $4,425. Since the population standard deviation is unknown, a t-test is appropriate. The significance level is α = 0.01, and the sample size is small.

The hypotheses are formulated as follows:

• Null hypothesis (H0): μ = $4,425
• Alternative hypothesis (Ha): μ ≠ $4,425

Because the focus is on whether the mean balance differs from $4,425, this is a two-tailed test. The t-statistic is calculated using the formula:

t = (x̄ - μ₀) / (s / √n)

where x̄ is the sample mean, μ₀ is the hypothesized mean, s is the sample standard deviation, and n is the sample size. Substituting the values yields:

t = (4300 - 4425) / (400 / √15) ≈ -1.56

Using a t-distribution table or software, the p-value associated with t = -1.56 with degrees of freedom (df = n-1 = 14) can be obtained. Suppose the p-value is approximately 0.14. Since p = 0.14 is greater than α = 0.01, we fail to reject the null hypothesis. This indicates there is no sufficient evidence to suggest the mean balance significantly differs from $4,425.

The conclusion: “At a 1% significance level, there is not enough evidence to reject the claim that the mean account balance is $4,425.”

Scenario 3: Home asking prices

A sample of 350 homes has a mean asking price of $247,000, with a population standard deviation of $8,000. The city claims the mean asking price is $253,000. The hypothesis test will determine whether the mean asking price is less than $253,000, using a significance level of α = 0.10.

Formulating hypotheses:

• Null hypothesis (H0): μ = $253,000
• Alternative hypothesis (Ha): μ

This is a one-tailed test because the question asks if the mean is less than the claimed value. The z-test is appropriate because the population standard deviation is known. The z-statistic is calculated as:

z = (x̄ - μ₀) / (σ / √n) = (247,000 - 253,000) / (8,000 / √350) ≈ -5.18

Consulting the standard normal distribution table or software, the p-value associated with a z-score of -5.18 is very close to zero (

Conclusion

Across all three scenarios, hypothesis testing allows researchers to make informed decisions based on sample data and the likelihood of observing such data if the null hypothesis were true. When applying these tests, selecting the appropriate test statistic (z or t), formulating correct hypotheses, computing accurate test statistics, and interpreting p-values in the context of the significance level are crucial steps. These procedures enable decision-makers such as professors, marketing managers, or city officials to assess claims and make data-driven conclusions about populations.

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