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In a natural population of mice (Mus musculus) near Ann Arbor, Michigan, the coats of some individuals are white spotted on the belly. In a sample of 580 mice from the population, 28 individuals were found to have white-spotted bellies. \({ }^{8}\) Construct a \(95 \%\) confidence interval for the population proportion of this trait.

Short Answer

Expert verified
The 95% confidence interval for the population proportion of white-spotted bellies in the mice population is \(\text{p-hat} - ME\) to \(\text{p-hat} + ME\).

Step by step solution

01

Identify the sample proportion

First, calculate the sample proportion (\( \text{p-hat} \)) of white-spotted bellies by dividing the number of individuals with the trait by the total number of individuals sampled. The formula to find the sample proportion is given by: \(\text{p-hat} = \frac{\text{number of successes}}{\text{total sample size}}\).
02

Calculate the standard error

Second, calculate the standard error of the sample proportion using the formula: \(\text{SE} = \sqrt{\frac{\text{p-hat}(1-\text{p-hat})}{n}}\), where \(n\) is the sample size.
03

Determine the Z-score for the confidence level

Look up the Z-score that corresponds to a 95% confidence level in a Z-score table or use a statistical software. The Z-score for a 95% confidence level is typically 1.96.
04

Calculate the margin of error

The margin of error (ME) is calculated using the formula: \(ME = Z \times \text{SE}\).
05

Construct the confidence interval

The confidence interval is found using the formula: \(\text{CI} = \text{p-hat} \pm ME\). Calculate the lower and upper bounds of the confidence interval using the sample proportion, margin of error, and the Z-score.

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Key Concepts

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

Sample Proportion
A crucial concept in statistics is the sample proportion, which represents the fraction of individuals in a sample that display a certain characteristic. In our exercise, we focused on mice with white-spotted bellies. To calculate the sample proportion (\text{p-hat}), we divide the number of mice with the trait (our successes) by the entire number of mice we've sampled. This gives us:
\[\text{p-hat} = \frac{28}{580}\]
From this calculation, we obtain an estimated proportion of the characteristic within the sample which then is used to infer the proportion in the entire population. It's important to note that the accuracy of our sample proportion depends on the sample size and randomness. Larger, more representative samples generally provide a proportion closer to the true population proportion.
Standard Error
Once we have the sample proportion, we proceed to calculate the standard error (SE), a measure of the variability or precision of the sample proportion. It’s calculated using:
\[\text{SE} = \sqrt{\frac{\text{p-hat}(1-\text{p-hat})}{n}}\]
For our mice example, we substitute \text{p-hat} and sample size (\text{n}) into the formula to determine how much the sample proportion might differ from the true population proportion due to sampling error. The standard error diminishes as the sample size increases, meaning larger samples give us a more precise estimate. Conversely, a high standard error implies less confidence in the precision of our sample proportion, highlighting the variability inherent in our sample.
Z-score
Lastly, the concept of the Z-score is pivotal when constructing confidence intervals. A Z-score is a statistical measure that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. For a 95% confidence level, the Z-score tells us how many standard errors we must add or subtract from \text{p-hat} to ensure that the interval captures the true population proportion 95% of the time. Typically, this Z-score is 1.96.
When we calculate the margin of error (ME) as \(ME = Z \times \text{SE}\), we apply the Z-score to our standard error, thus accounting for the variability in our estimate. This step adjusts the width of the confidence interval, directly impacting how certain we can be that our interval encompasses the true population parameter. The Z-score is a bridge between our sample data and the level of certainty we seek to establish in our conclusions about the population.

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Most popular questions from this chapter

Geneticists studying the inheritance pattern of cowpea plants classified the plants in one experiment according to the nature of their leaves. The data follow : $$ \begin{array}{|lccc|} \hline \text { Type } & \text { I } & \text { II } & \text { III } \\ \hline \text { Number } & 179 & 44 & 23 \\ \hline \end{array} $$ Test the null hypothesis that the three types occur with probabilities \(12 / 16,3 / 16,\) and \(1 / 16 .\) Use a chi-square test with \(\alpha=0.10\).

An experiment was conducted in which two types of acorn squash were crossed. According to a genetic model, \(1 / 2\) of the resulting plants should have dark stems and dark fruit, \(1 / 4\) should have light stems and light fruit, and \(1 / 4\) should have light stems and plain fruit. The actual data were \(220,129,\) and 105 for these three categories. \({ }^{49}\) Do these data refute this model? Consider a chi-square test. (a) What is the value of the chi-square test statistic for investigating whether these data are consistent with the \(1 / 2,1 / 4,1 / 4\) probabilities model? (b) The \(P\) -value for the chi-square test is \(0.23 .\) If \(\alpha=0.10\), what is your conclusion regarding \(H_{0} ?\)

In a study of resistance to a certain soybean virus, biologists cross- fertilized two soybean cultivars. They expected to get a 3: 1 ratio of resistant to susceptible plants. The observed data were 58 resistant and 26 suscep- tible plants. \(^{50}\) Are these data significantly inconsistent with the expected 3: 1 ratio? Consider a test, using \(\alpha=0.10\) and a nondirectional alternative. (a) What are the expected counts for the two categories under the null hypothesis? (b) The \(P\) -value for the chi-square test (with a nondirectional alternative) is \(0.21 .\) If \(\alpha=0.10,\) what is your conclusion regarding \(H_{0} ?\) (c) Do these results confirm the 3: 1 ratio expected by the researchers?

Among \(n\) babies born in a certain city, \(51 \%\) were boys. Suppose we want to test the hypothesis that the true probability of a boy is \(\frac{1}{2} .\) Calculate the value of \(\chi_{s}^{2}\) and bracket the \(P\) -value for testing against a nondirectional alternative, if (a) \(n=1,000\) (b) \(n=5,000\) (c) \(n=10,000\)

Gene mutations have been found in patients with muscular dystrophy. In one study, it was found that there were defects in the gene coding of sarcoglycan proteins in 23 of 180 patients with limb-girdle muscular dystrophy. \({ }^{19}\) Use these data to construct a \(99 \%\) confidence interval for the corresponding population proportion.

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