The Sugar Content Of Canned Peaches Syrup
The Sugar Content Of The Syrup In Canned Peaches Is Normally Distrib
The sugar content of the syrup in canned peaches is normally distributed, and the variance is thought to be 218 (mg)^2. The problem involves testing the hypothesis regarding this variance, analyzing the data using the appropriate statistical approaches, constructing confidence intervals, and verifying assumptions related to normality and equal variances. Specifically, you are asked to conduct a hypothesis test for the variance, determine the P-value, construct a confidence interval for the variance, and interpret these results. Additionally, there is a part that involves constructing normal probability plots to assess the normality assumption, performing a two-sample test for mean differences, and constructing confidence intervals for the difference in means. The primary focus here is on understanding and applying hypothesis testing concepts, confidence intervals, and the assessment of underlying assumptions in the context of normally distributed data.
Paper For Above instruction
The evaluation of whether the variance of the sugar content in canned peaches’ syrup differs from a specified value involves a hypothesis test rooted in the chi-square distribution, which is appropriate for variance testing in normally distributed data. Given the sample data—a random sample of n=10 cans with a sample standard deviation of s=4 mg—the null hypothesis postulates that the population variance is 18 (mg)^2, while the alternative hypothesis suggests that it is not equal to this value. This analytical process necessitates several structured steps, including the formulation of hypotheses, calculation of the test statistic, determination of the rejection region, and decision-making based on the significance level (α=0.05).
Assumptions and Hypotheses
The critical assumption underlying this test is that the data are normally distributed, which justifies the use of chi-square distributions for variance testing. The hypotheses are set as follows:
- Null hypothesis (H₀): σ² = 18 (mg)^2
- Alternative hypothesis (H₁): σ² ≠ 18 (mg)^2
Step-by-step Rejection Region Approach
1. State the hypotheses (as above) and the significance level.
2. Calculate the test statistic:
\[
\chi^2 = \frac{(n-1)s^2}{σ_0^2} = \frac{(10-1)(4)^2}{18} = \frac{9 \times 16}{18} = \frac{144}{18} = 8
\]
3. Determine the degrees of freedom: df = n−1 = 9.
4. Find the critical chi-square values at α=0.05 for a two-tailed test:
\[
\chi^2_{0.025,9} \approx 19.023,\quad \chi^2_{0.975,9} \approx 2.700
\]
5. Rejection region:
- Reject H₀ if \(\chi^2 19.023\).
6. Decision: Since the calculated χ² (8) is between 2.700 and 19.023, we do not reject H₀ at α=0.05.
Part (b): P-value Calculation
Using the chi-square distribution with 9 degrees of freedom, the P-value corresponds to the probability under the chi-square curve for a test statistic as extreme or more extreme than the one observed. The two-sided P-value is calculated as:
\[
\text{P-value} = 2 \times \min \left( P(\chi^2 \leq 8), P(\chi^2 \geq 8) \right)
\]
Consulting chi-square tables or software:
- \(P(\chi^2 \leq 8)\) ≈ 0.440
- Therefore, P-value ≈ 2 × 0.440 = 0.880.
Since this P-value is much larger than 0.05, the evidence is weak to reject H₀.
Part (c): 95% Confidence Interval for Variance
The confidence interval for the population variance is given by:
\[
\left[\frac{(n−1)s^2}{\chi^2_{1−\alpha/2,\,n−1}}, \frac{(n−1)s^2}{\chi^2_{\alpha/2,\,n−1}}\right]
\]
Substituting the values:
\[
\left[\frac{9 \times 16}{\chi^2_{0.975,9}},\, \frac{9 \times 16}{\chi^2_{0.025,9}}\right] = \left[\frac{144}{2.700},\, \frac{144}{19.023}\right] \approx [53.33, 7.58]
\]
Since the variance is a squared quantity, the interval indicates the plausible range of the population variance, which includes 18, supporting the earlier conclusion.
Part (d): Using the Confidence Interval to Test the Hypothesis
Because 18 falls well within the calculated interval (7.58 to 53.33), it suggests that the hypothesized variance of 18 is plausible at the 95% confidence level. This aligns with the earlier conclusion that the data do not provide sufficient evidence to reject the null hypothesis.
Assessing Normality and Variance Assumptions
Constructing normal probability plots (Q-Q plots) for the sample data visually assesses the normality assumption. If the points approximately follow a straight line, this supports normality. In practical applications, tools like Minitab can generate these plots, which should be examined for deviations such as curvature or outliers. If the plots show no substantial deviations, the assumption of normality is reasonable.
Similarly, equal variances between different samples can be assessed via residual plots or Levene’s or Bartlett’s tests, supporting the assumption of homogeneity of variances necessary for certain parametric tests.
Comparison of Mean Deflection Temperatures
In the context of comparing two types of pipe, similar hypothesis testing procedures can be applied, typically involving two-sample t-tests. The P-value approach with α = 0.05 evaluates whether the observed difference is statistically significant, while the critical region approach confirms whether the test statistic falls into the rejection region. Confidence intervals for mean differences further quantify the magnitude and direction of the differences and can reinforce the findings from hypothesis testing when the interval excludes zero.
Conclusion
The statistical analysis applied to the variability of sugar content in canned peaches suggests that the observed sample does not provide sufficient evidence to conclude that the population variance differs from 18 (mg)^2 at the 5% significance level. The confidence intervals and P-values corroborate this conclusion, emphasizing the importance of considering assumptions and the use of multiple methods for inference. Normal probability plots and tests for equality of variances are essential in validating the appropriateness of the parametric tests used.
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