Name Module 3 Homework Assignment ✓ Solved

Name Module 3 Homework Assignmen

Determine whether the hypothesis test involves a sampling distribution of means that is a normal distribution, Student t distribution or neither. The sample data appear to come from a normally distributed population with σ = 28. Explain. Claim: μ = 977. Sample data: n = 25, = 984, s = 25. Explain.

In a sample of 100 M&M’s, it is found that 8% are brown. Use a 0.05 significance level to test the claim that the percentage of brown M&M’s is equal to 13%.

Identify the null and alternative hypotheses.

(Refer to problem 2) Find the test statistic.

(Refer to problem 2) Use the Standard Normal Table to find the p-value, critical value(s) and state the decision about the null hypothesis.

(Refer to problem 2) State the conclusion in non-technical terms.

Test the claim that for the adult population of one town, the mean annual salary is given by μ = 30,000. Sample data are summarized as n = 17, = $22,298, and s = $14,200. Use a significance level of α = 0.05. State the null and alternative hypotheses.

(Refer to Question 6) Find the test statistic.

(Refer to Question 6) Find the critical value(s) and state decision about null hypothesis.

(Refer to Question 6) State the conclusion in non-technical terms.

Find the critical value or values of χ² based on the given information. H₁: σ > 26.1, n = 9, α = 0.01

Sample Paper For Above instruction

Hypothesis testing is a fundamental aspect of inferential statistics, allowing researchers to make decisions about population parameters based on sample data. In this paper, we will explore several hypothesis testing scenarios, emphasizing the determination of the appropriate sampling distribution, formulation of hypotheses, calculation of test statistics, interpretation of p-values, critical values, and the conclusions drawn in both statistical and layman's terms.

Assessing the Distribution Type in a Mean Population

The first scenario involves determining whether a hypothesis test concerning the population mean employs a normal distribution, Student's t distribution, or neither. The sample originates from a normally distributed population with a known standard deviation (σ = 28). The sample size is 25, with a sample mean ( = 984) and sample standard deviation (s = 25). Since the population standard deviation σ is known, and the sample size exceeds 30, the sampling distribution of the mean follows a normal distribution according to the Central Limit Theorem. Specifically, when σ is known and the sample size is large (n ≥ 30), the sampling distribution of the sample mean is approximately normal. Thus, in this case, the normal distribution is appropriate for hypothesis testing about the population mean. If σ were unknown and the sample size comparatively small, the Student's t distribution would be used instead (since the sample appears to come from a normally distributed population). Therefore, given the known σ and sufficient sample size, the hypothesis test involves a normal distribution.

Testing Population Proportion for M&Ms

The second scenario involves testing a claim about the proportion of brown M&Ms in a sample. A sample of 100 M&Ms shows that 8% are brown, and the significance level is set at 0.05. The null hypothesis (H₀) posits that the true proportion of brown M&Ms is 13% (p₀ = 0.13), while the alternative hypothesis (H₁) suggests that the true proportion differs from 13%. To test this, the sample proportion (p̂) is 0.08. Since the sample size is large, z-tests for proportions are appropriate. The test statistic is calculated using the formula:

Z = (p̂ - p₀) / sqrt[ p₀(1 - p₀) / n ].

Substituting the values:

Z = (0.08 - 0.13) / sqrt[ 0.13(0.87) / 100 ] ≈ -2.72.

Using the standard normal distribution table, the p-value associated with Z ≈ -2.72 is about 0.0065, which is less than the significance level (0.05). The critical z-value for a two-tailed test at α = 0.05 is approximately ±1.96. Since |Z| > 1.96, we reject the null hypothesis. Therefore, there is sufficient evidence to conclude that the proportion of brown M&Ms differs from 13% at the 5% significance level. In lay terms, the data suggests that the true percentage of brown M&Ms is not 13%, but rather significantly different, likely lower in this sample.

Testing the Population Mean Salary

The third scenario pertains to testing the mean annual salary of adults in a town, where the claimed mean is $30,000. The sample size is 17, with a mean salary of $22,298 and a standard deviation of $14,200. The significance level is 0.05. Since the population standard deviation is unknown and the sample size is small (

t = (̄x - μ₀) / (s / sqrt(n)).

Plugging in the values:

t = (22,298 - 30,000) / (14,200 / sqrt(17)) ≈ -1.322.

Degrees of freedom = n - 1 = 16. The critical t-value at α = 0.05 for a two-tailed test is approximately ±2.12. Since |t| ≈ 1.322

Testing Variance Using Chi-Square

The final scenario involves testing the variance of salaries in a town, with a claim that the population standard deviation exceeds 26.1 based on a sample of 9 individuals. The sample variance (s²) is derived from the sample standard deviation, and the test employs the chi-square distribution due to its focus on variance. The null hypothesis H₀: σ² ≤ 26.1², and the alternative hypothesis H₁: σ² > 26.1². The test statistic is calculated by:

χ² = (n - 1) * s² / σ₀².

Using the sample data, the critical value of χ² at α = 0.01 with 8 degrees of freedom is obtained from chi-square tables. Since the alternative hypothesis indicates a right-tailed test, if the computed χ² exceeds the critical value, we reject H₀, supporting the claim that the population variance is greater than the hypothesized level. This test helps in understanding the variability within the population; in this context, a higher variance supports greater unpredictability in salaries or other measures analyzed.

Conclusion

Throughout these scenarios, the importance of selecting the correct distribution and understanding the parameters involved in hypothesis testing is evident. Whether dealing with means, proportions, or variances, the choice of the test—normal, t, or chi-square—depends on known parameters, sample size, and the nature of the data. Interpreting results in both statistical and non-technical terms ensures that findings are accessible and meaningful to broader audiences. Proper application of these techniques is essential for accurate inference and data-driven decision-making in diverse fields such as quality control, social sciences, business analytics, and research.

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