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In \(1964,40 \%\) of shipments of U.S. cotton to Germany were sent to arbitration because of complaints about the quality being substandard, according to Time Magazine. Would it signify a real worsening of the situation if 20 out of the first 40 shipments in 1965 were likewise the cause of complaint, or might this difference be attributed to chance?

Short Answer

Expert verified
The proportions of complaints in 1964 and 1965 were 40% and 50%, respectively. Using a proportion comparison test, the calculated z-score was approximately 1.49. With a 95% confidence level, the critical value for a two-tailed test is approximately ±1.96. Since the calculated z-score falls within the critical value range, the difference in the proportions of complaints between the two years can be attributed to chance and does not signify a real worsening of the situation.

Step by step solution

01

Determine Proportions

First, calculate the proportions for comparison: 1964: \(\frac{40}{100} = 0.40\) 1965: \(\frac{20}{40} = 0.50\) We want to know if the proportion of complaints in 1965 (0.50) is significantly different from the proportion in 1964 (0.40).
02

Calculate Combined Proportion

Combine the two proportions to calculate the overall proportion (p): \[p = \frac{(n_1 *p_1) + (n_2 * p_2)}{n_1 + n_2}\] Where \(n_1\) and \(n_2\) are the sample sizes and \(p_1\) and \(p_2\) are the proportions. In this case, \(n_1 = n_2 = 40\), \(p_1 = 0.40\), and \(p_2 = 0.50\). \[p = \frac{(40 * 0.40) + (40 * 0.50)}{40 + 40}\] \[p = \frac{16 + 20}{80}\] \[p = \frac{36}{80} \Rightarrow p = 0.45\]
03

Calculate the Z-Score

Calculate the z-score: \[z = \frac{(p_2 - p_1) - 0}{\sqrt{p(1-p)(\frac{1}{n_1}+\frac{1}{n_2})}}\] Now plug in the values: \[z = \frac{0.50 - 0.40}{\sqrt{(0.45)(1-0.45)(\frac{1}{40}+\frac{1}{40})}}\] \[z \approx \frac{0.10}{\sqrt{(0.45)(0.55)(\frac{1}{20})}}\] \[z \approx 1.49\]
04

Determine the Critical Value and Draw Conclusion

Assuming a 95% confidence level, the critical value for a two-tailed test is approximately ±1.96. The z-score calculated (1.49) does not exceed the critical value of ±1.96. Since 1.49 falls within the critical value range, the difference in the proportions of complaints between the two years can be attributed to chance and does not signify a real worsening of the situation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proportions
Proportions in statistics represent a part of a whole and are often expressed as a fraction or percentage. They help us understand the proportion of interest in a dataset. In hypothesis testing, proportions can tell us if there’s a significant change or difference in sample proportions over time or between groups.
In the given exercise, the proportion of U.S. cotton shipments to Germany that were sent for arbitration was compared between two years, 1964 and 1965. We calculated the proportion for 1964 as 0.40, reflecting 40%, and for 1965 as 0.50, reflecting 50%. This comparison is to determine if the increase in complaints in 1965 might signify a real issue or if it could just be due to chance.
To make comparisons easier, we often calculate a combined proportion, which helps in establishing a uniform proportion level that represents the expected complaint rate if the difference was purely by chance. This is crucial for calculating further statistical metrics such as the z-score, which will test the significance of the observed difference.
Z-Score
The z-score is a statistical metric used to determine the number of standard deviations an element is from the mean. It's a fundamental part of hypothesis testing, as it helps in comparing a sample proportion to a population proportion or another sample proportion.
In this exercise, the z-score was calculated to examine how far the 1965 proportion of complaints deviates from the 1964 proportion. By using the formula:\[ z = \frac{(p_2 - p_1) - 0}{\sqrt{p(1-p)(\frac{1}{n_1}+\frac{1}{n_2})}} \]where \(p_1\) and \(p_2\) refer to the proportions of the different years, and \(n_1\) and \(n_2\) represent the sample sizes, the z-score tells us whether the observed differences could likely be due to statistical variation alone.
In this scenario, the computed z-score came out to approximately 1.49. This value will now be used to compare against the critical values to make an informed decision about the significance of the difference observed.
Critical Value
Critical values are threshold values that define regions where your observed test statistic would lead you to reject the null hypothesis. These values are closely related to confidence levels in hypothesis testing.
To determine if the difference in proportions between 1964 and 1965 is significant, we compare the calculated z-score against the critical value. For a two-tailed test, with a 95% confidence level, the critical values are ±1.96. This means any z-score beyond these thresholds would suggest the difference is statistically significant.
Here, the calculated z-score was 1.49, which lies within the interval of -1.96 to +1.96. Hence, it indicates that the observed difference in proportions does not exceed the critical value range. This suggests that the observed differences might be due to random chance rather than a true deterioration of conditions. The critical value acts as a benchmark, guiding whether the findings in the sample can be generalized to the population.
Significance Level
The significance level, often denoted by α, is a probabilistic metric that helps in hypothesis testing. It represents the risk of concluding that a difference exists when there is none—a Type I error. Commonly set at 0.05 (5%), this level provides a cut-off for decision-making.
In the exercise, a 95% confidence level was assumed, equivalent to a significance level of 0.05, reflecting that there is 5% risk of wrongly claiming that the proportions of complaints in the shipments have significantly increased.
When conducting a hypothesis test, this significance level determines the critical value thresholds. If the z-score exceeds these thresholds, the findings are statistically significant. Otherwise, as in this scenario, when the z-score of 1.49 falls within the critical range, it suggests that there is no statistically significant change, affirming that variations might be due to random sampling variability.
Understanding the significance level allows us to judge the reliability of our statistical conclusions. It outlines how confident we are in our data representations and decisions.

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Most popular questions from this chapter

A manufacturer of car batteries guarantees that his batteries will last, on the average, 3 years with a standard deviation of 1 year. If 5 of these batteries have lifetimes of \(1.9,2.4,3.0\), \(3.5\), and \(4.2\) years, is the manufacturer still convinced that his batteries have a standard deviation of 1 year?

In a recent science fiction novel, aliens invented a death ray that killed 3 Earthman. The lethal doses were 10,11 , and 12 units of the ray. The mean lethal dose for Moonmen is \(13.30\) units. For this sample of 3 Earthmen \(\underline{\mathrm{X}}=11\) and \(\mathrm{S}=1.0\) and a test of significance of the difference between the mean for Earthmen and Moonmen yields \(\mathrm{t}=3.98\), not significant at the \(\alpha=.05\) level of significance. Then 6 more Earthmen were killed with lethal doses of \(9,10,11,12,13\) and \(14 .\) For this sample, \(\underline{\mathrm{X}}=11.5\), and \(\mathrm{S}=1.87\), and a test of significance of the difference between the mean for Earthmen and Moonmen yields \(\mathrm{t}=2.36\), also not significant at \(\alpha=.05\). What can be said if we combine the results of these two samples?

Suppose it is required that the mean operating life of size "D" batteries be 22 hours. Suppose also that the operating life of the batteries is normally distributed. It is known that the standard deviation of the operating life of all such batteries produced is \(3.0\) hours. If a sample of 9 batteries has a mean operating life of 20 hours, can we conclude that the mean operating life of size "D" batteries is not 22 hours? Then suppose the standard deviation of the operating life of all such batteries is not known but that for the sample of 9 batteries the standard deviation is \(3.0 .\) What conclusion would we then reach?

A consumer report shows that in testing 8 tires of brand \(\mathrm{A}\) the mean life of the tires was 20,000 miles with a standard deviation of 2,500 miles. Twelve tires of brand \(B\) were tested under similar conditions with a mean life of 23,000 miles and a standard deviation of 2,800 miles. If a .05 level of significance is used, does the data present sufficient evidence to indicate a difference in the average life of the two brands of tires?

An experimenter produced two kinds of lesions in monkeys Lesion 1 in one group, Lesion 2 in another group. Then he observed each animal for a month after surgery and rated the amount of aggression each animal displayed. He theorized that the two groups would differ primarily in variability. Here is the data: \(\begin{array}{ll}\text { Lesion 1 } & \text { Lesion } 2\end{array}\) $$ \begin{array}{rlrl} \underline{\mathrm{X}} & =95.3 & & \underline{\mathrm{X}}=94.8 \\ \mathrm{~s}^{2} & =13.7 & \mathrm{~s}^{2}=6.13 \\ \mathrm{~N} & =31 & \mathrm{~N}=31 \end{array} $$ Do the groups differ in variability? Assume that the scores are normal, and use the \(.02\) level of significance.

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