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Dolly the Sheep, the world's first mammal to be cloned, was introduced to the public in 1997. In a Pew Research poll taken soon after Dolly's debut, \(63 \%\) of Americans were opposed to the cloning of animals. In a Pew Research poll taken 20 years after Dolly, \(60 \%\) of those surveyed were opposed to animal cloning. Assume this was based on a random sample of 1100 Americans. Does this survey indicate that opposition to animal cloning has declined since \(1997 ?\) Use a \(0.05\) significance level.

Short Answer

Expert verified
Yes, the survey indicates that opposition to animal cloning among Americans has statistically significantly declined since 1997.

Step by step solution

01

Define the hypotheses

The null hypothesis (\(H_0\)) is that the proportion of those opposed to animal cloning has not changed, and the alternative hypothesis (\(H_1\)) is that the proportion of those opposed has decreased since 1997. Formally, these are: \(H_0: p_{1997} = p_{2017}\) and \(H_1: p_{1997} > p_{2017}\).
02

Calculate the sample proportions

The sample proportions (\(p\)) are the percentages given in the problem. Hence, \(p_{1997} = 0.63\) and \(p_{2017} = 0.60\). The combined sample proportion (\(p\)) is \((0.63*1100 + 0.60*1100) / (2*1100) = 0.615\).
03

Calculate the standard error

The formula for the standard error for two sample proportions is \(\sqrt{ p(1 - p) * [(1/n_1) + (1/n_2)] }\). Substituting the values in, we get \(SE = \sqrt{0.615 * (1 - 0.615) * [(1/1100) + (1/1100)]} = 0.0158\).
04

Calculate the z-score

The z-score is calculated using the formula: \(z = (p_{1} - p_{2}) / SE\), where \(p_{1}\) and \(p_{2}\) are the sample proportions, and SE is the standard error. Calculating, we get \(z = (0.63 - 0.60) / 0.0158 = 1.899\). Refer to the standard normal (Z-) table to find the corresponding p-value.
05

Compare p-value with significance level

The corresponding p-value for a z-score of 1.899 is approximately 0.028. This p-value is less than the significance level of 0.05, so we reject the null hypothesis. This means that there is a statistically significant decrease in opposition to animal cloning since 1997.

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Key Concepts

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

Significance Level
The significance level, often denoted as \( \alpha \), is a critical threshold in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true. This concept is key in determining the strength of our evidence against the null hypothesis. In practical terms, the significance level is the cut-off for determining when the observations are unlikely enough under the null hypothesis that we might consider the alternative hypothesis. For instance, if we choose a significance level of 0.05, it implies that there's a 5% risk of concluding that a difference exists when there is, in fact, no difference. Here's what you need to keep in mind about significance level:
  • A common choice for significance level is 0.05, but this can vary depending on the context and field of study.
  • The lower the significance level, the stronger the evidence required to reject the null hypothesis.
  • Significance level is pre-chosen before conducting the test to stay unbiased.
In this exercise, with a significance level of 0.05, the decision is whether evidence from the data provides sufficient grounds to believe in a decrease in opposition to animal cloning. A p-value less than this level prompts us to reject the null hypothesis.
Sample Proportions
Sample proportions are used to estimate the population proportion. In hypothesis testing, the sample proportion is a key element as it provides insight into the characteristics of the larger population. In this exercise, the sample proportions reflect the percentage of individuals opposed to animal cloning in the years 1997 and 2017, which are 0.63 and 0.60 respectively. These values are computed based on the responses of 1100 Americans, which is a substantial sample size. Some crucial points about sample proportions include:
  • The sample proportion \( p \) is calculated by dividing the number of favorable responses by the total sample size.
  • A larger sample size tends to produce more reliable estimates of the population proportion.
  • It's important to ensure that the sample is random to avoid bias in the results.
Using sample proportions, the exercise explores how the public opinion on animal cloning may have shifted over two decades. Calculating these proportions accurately is vital, as they are used in further statistical analysis such as computing the standard error and z-score.
Z-score Calculation
The z-score is an important statistic in hypothesis testing as it helps identify how far away a sample proportion is from the population proportion, in terms of standard errors. It serves as a bridge in understanding if the observed data significantly deviates from what was expected under the null hypothesis.In this scenario, the z-score is used to determine if the change from a 63% opposition in 1997 to a 60% opposition in 2017 is a significant drop. Here's how to compute and interpret the z-score in this context:
  • The z-score formula is: \( z = \frac{p_1 - p_2}{SE} \), where \( p_1 \) and \( p_2 \) are the sample proportions, and \( SE \) is the standard error.
  • Using these values, the calculated z-score is 1.899. This measures how many standard errors the sample proportion from 2017 is below the 1997 proportion.
  • A higher absolute value of the z-score can indicate a more significant difference between the two sample proportions.
The resultant z-score of 1.899 is then used to look up the p-value in a standard normal distribution table. A smaller p-value compared to the significance level suggests that the observed change is statistically significant, leading to potential rejection of the null hypothesis. In summary, the z-score is essential for gauging the significance of the observed trends in survey responses.

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Most popular questions from this chapter

A community college used enrollment records of all students and reported that that the percentage of the student population identifying as female in 2010 was \(54 \%\) whereas the proportion identifying as female in 2018 was \(52 \%\). Would it be appropriate to use this information for a hypothesis test to determine if the proportion of students identifying as female at this college had declined? Explain.

Votes for Independents Judging on the basis of experience, a politician claims that \(50 \%\) of voters in Pennsylvania have voted for an independent candidate in past elections. Suppose you surveyed 20 randomly selected people in Pennsylvania, and 12 of them reported having voted for an independent candidate. The null hypothesis is that the overall proportion of voters in Pennsylvania that have voted for an independent candidate is \(50 \%\). What value of the test statistic should you report?

Morse determined that the percentage of \(a\) 's in the English language in the 1800 s was \(8 \%\). A random sample of 600 letters from a current newspaper contained 60 a's. Using the \(0.10\) level of significance, test the hypothesis that the proportion of \(a\) 's in this modern newspaper is \(0.09\).

According to a 2016 report from the Institute for College Access and Success \(66 \%\) of all graduates from public colleges and universities had student loans. A public college surveyed a random sample of 400 graduates and found that \(62 \%\) had student loans. a. Test the hypothesis that the percentage of graduates with student loans from this college is different from the national percentage. Use a significance level of \(0.05\). b. After conducting the hypothesis test, a further question one might ask is what proportion of graduates from this college have student loans? Use the sample data to find a \(95 \%\) confidence interval for the proportion of graduates from the college who have student loans. How does this confidence interval support the hypothesis test conclusion?

In 2015 a Gallup poll reported that \(52 \%\) of Americans were satisfied with the quality of the environment. In 2018 , a survey of 1024 Americans found that 461 were satisfied with the quality of the environment. Does this survey provide evidence that satisfaction with the quality of the environment among Americans has decreased? Use a \(0.05\) significance level.

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