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91Ó°ÊÓ

In one income group, \(45 \%\) of a random sample of people express approval of a product. In another income group, \(55 \%\) of a random sample of people express approval. The standard errors for these percentages are \(.04\) and \(.03\) respectively. Test at the \(10 \%\) level of significance the hypothesis that the percentage of people in the second income group expressing approval of the product exceed that for the first income group.

Short Answer

Expert verified
In conclusion, a hypothesis test was performed at a 10% significance level, and the results provided sufficient evidence to reject the null hypothesis, supporting the alternative hypothesis that the percentage of people in the second income group expressing approval of the product exceeds that for the first income group. The test statistic (z) was 2 and the critical value (z-critical) was 1.28. Since \( z > z_{critical} \) (2 > 1.28), we reject the null hypothesis in favor of the alternative hypothesis.

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis (H0) states that there is no significant difference between the approval percentages of the two income groups, which means that the difference between the approval percentages is equal to or less than 0. The alternative hypothesis (H1) states that the approval percentage in the second income group is greater than that of the first income group. Mathematically, we can represent the hypotheses as follows: - H0: p2 ≤ p1 - H1: p2 > p1
02

Calculate the test statistic (z-score)

The test statistic for comparing two proportions is the z-score, which is calculated as follows: \[z = \frac{(p_2-p_1) - 0}{\sqrt{SE_1^2 + SE_2^2}}\] where p1 and p2 are the proportions, and SE1 and SE2 are the standard errors for the two income groups. In our case, p1 = 0.45, p2 = 0.55, SE1 = 0.04, and SE2 = 0.03. Substituting the values, we get: \[z = \frac{(0.55-0.45) - 0}{\sqrt{0.04^2 + 0.03^2}}\]
03

Compute the z-score

\[z = \frac{0.10}{\sqrt{0.0016 + 0.0009}}\] \[z = \frac{0.10}{\sqrt{0.0025}}\] \[z = \frac{0.10}{0.05}\] \[z = 2\]
04

Determine the critical value and make a decision

Since we are testing at a 10% level of significance for a one-tailed test, the critical value (z-critical) can be found in the z-table or using a calculator. The critical value for a one-tailed test at a 10% significance level is 1.28. Now, compare the test statistic (z) with the critical value (z-critical): - If z > z-critical, then we reject the null hypothesis. - If z ≤ z-critical, then we fail to reject the null hypothesis. In our case, the test statistic z = 2 and the critical value z-critical = 1.28. Since z > z-critical (2 > 1.28), we reject the null hypothesis.
05

Conclusion

Based on the hypothesis test, we can conclude that there is sufficient evidence at the 10% significance level to support the alternative hypothesis that the percentage of people in the second income group expressing approval of the product exceeds that for the first income group.

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Key Concepts

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

Null and Alternative Hypotheses
In hypothesis testing, we start by establishing two competing statements: the null hypothesis and the alternative hypothesis. The null hypothesis (H_0) is a statement of no effect, no difference, or a scientific hypothesis that a researcher attempts to disprove. For instance, if we have two income groups and we are investigating whether their approval for a product is the same, the null hypothesis would state that there is no significant difference between the approval percentages.The alternative hypothesis (H_1 or H_a), on the other hand, represents a statement contrary to the null hypothesis. It's what you might choose to believe if you reject the null hypothesis. In the same scenario, this could be the supposition that one income group has a higher approval percentage for a product than the other. It is essential to define these hypotheses precisely because they establish the basis of the testing procedure.
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's used to determine the likelihood that the null hypothesis is true. In many cases, the test statistic follows a certain probability distribution under the assumption that the null hypothesis is correct. For comparing two proportions, as in the exercise, we often use the Z-test to compute a Z-score, which represents how many standard deviations a data point is from the mean.A/z-test statistic is usually calculated by taking the difference between the sample statistic and the null hypothesis, then dividing it by the standard error. The process is different depending on the type of data and the test being performed, but the aim is to quantify the weight of evidence against the null hypothesis.
Z-score
A Z-score is a numerical measurement used in statistics to describe a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. If a Z-score is 0, it indicates that the data point's score is identical to the mean score. A Z-score can be positive or negative, indicating the number of standard deviations it is above or below the mean, respectively. In hypothesis testing, the calculated Z-score is used as the test statistic for Z-tests. It allows us to determine how extreme or rare our data is under the null hypothesis and aids in deciding whether or not to reject the null hypothesis. High absolute values of Z-scores (both positive and negative) are associated with low probabilities of occurrence under the null hypothesis, which can be pivotal in the decision-making process of the test.
Level of Significance
The level of significance (\(alpha\)), often represented as a percentage, is a critical part of hypothesis testing. It defines the threshold for how extreme the test statistic must be for us to reject the null hypothesis. It's essentially the probability of making the wrong decision when the null hypothesis is actually true (Type I error).In practice, common choices for \(alpha\) include 0.05 (5%), 0.01 (1%), and 0.10 (10%). The level you choose can depend on the field of study or how much risk you're willing to accept of incorrectly rejecting the null hypothesis. The lower the significance level, the stronger the evidence must be to reject the null hypothesis.
Critical Value
The critical value is a point on the test distribution that is compared to the test statistic to decide whether to reject the null hypothesis. If your test statistic is more extreme than the critical value, then you reject the null hypothesis.The critical value depends on the predetermined level of significance and the distribution that applies to the test statistic. For Z-tests, the critical values are derived from the standard normal distribution. In the exercise, the test was one-tailed and the significance level was 10%, which meant the critical value was 1.28 (looking at Z-distribution tables or using statistical software). Since our calculated Z-score was greater than the critical value, it led to the rejection of the null hypothesis and supported the alternative hypothesis.

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Most popular questions from this chapter

It is believed that no more than \(40 \%\) of the students in a certain college wear glasses. Of 64 students surveyed, 40 or \(62.5 \%\) are found to wear glasses. Does this contradict the belief that no more than \(40 \%\) of the students wear glasses? Use a \(5 \%\) level of sionificance.

An electrical company claimed that at least \(95 \%\) of the parts which they supplied on a government contract conformed to specifications. A sample of 400 parts was tested, and 45 did not meet specifications. Can we accept the company's claim at a \(.05\) level of significance?

A certain brand of cigarettes is advertised by the manufacturer as having a mean nicotine content of 15 milligrams per cigarette. A sample of 200 cigarettes is tested by an independent research laboratory and found to have an average of \(16.2\) milligrams of nicotine content and a standard deviation of \(3.6\). Using a \(.01\) level of significance, can we conclude based on this sample that the actual mean nicotine content of this brand of cigarettes is greater than 15 milligrams?

Suppose the previous problem's experiment was carried out with a sample size of 8 . The mean, \(\underline{\mathrm{X}}\), for the 8 test cases for polished steel was 86,375 . The mean, \(\underline{Y}\), for the 8 test cases for unpolished steel was 79,838 . Can we conclude at a \(1 \%\) level of significance that the average endurance limit of polished specimens is greater than the average endurance limit for unpolished specimens?

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?

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