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USA Today reported that the state with the longest mean life span is Hawaii, where the population mean life span is 77 years. A random sample of 20 obituary notices in the Honolulu Advertizer gave the following information about life span (in years) of Honolulu residents: \(\begin{array}{llllllllll} 72 & 68 & 81 & 93 & 56 & 19 & 78 & 94 & 83 & 84 \\ 77 & 69 & 85 & 97 & 75 & 71 & 86 & 47 & 66 & 27 \end{array}\) i. Use a calculator with mean and standard deviation keys to verify that \(\bar{x}=\) \(71.4\) years and \(s\) \& \(20.65\) years. ii. Assuming that life span in Honolulu is approximately normally distributed, does this information indicate that the population mean life span for Honolulu residents is less than 77 years? Use a \(5 \%\) level of significance.

Short Answer

Expert verified
The sample does not provide enough evidence to show that the mean life span in Honolulu is less than 77 years (\(p > 0.05\)).

Step by step solution

01

State the Hypotheses

We need to determine whether the mean life span in Honolulu is less than 77 years. We set up our null hypothesis \( H_0: \mu = 77 \) (the population mean life span is 77 years) and our alternative hypothesis \( H_a: \mu < 77 \) (the population mean life span is less than 77 years).
02

Identify the Test and Critical Value

We will conduct a one-sample t-test for the population mean, as the sample size is 20 and the population standard deviation is unknown. With a significance level of \( \alpha = 0.05 \) and \( n-1 = 19 \) degrees of freedom, use a t-distribution table to find the critical value for a one-tailed test. The critical t-value is approximately -1.729.
03

Calculate the Test Statistic

Use the formula for the t-test statistic:\[t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\]Substitute \(\bar{x} = 71.4\), \(\mu = 77\), \(s = 20.65\), and \(n = 20\) into the formula:\[t = \frac{71.4 - 77}{20.65/\sqrt{20}} \approx \frac{-5.6}{4.619} \approx -1.213\]
04

Make a Decision

Compare the calculated test statistic \(-1.213\) with the critical t-value \(-1.729\). Since \(-1.213 > -1.729\), we fail to reject the null hypothesis. There is insufficient evidence to conclude that the population mean life span in Honolulu is less than 77 years at the \(5\%\) significance level.

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Key Concepts

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

t-test
A t-test is a type of statistical test used to compare the mean of a sample to a known value, which is often the mean of the overall population. This type of test is especially useful when working with small sample sizes and when the population standard deviation is unknown. In the original exercise, a one-sample t-test was employed to determine whether the average lifespan in Honolulu is different from 77 years. The small sample size of 20 justified using a t-test.

When conducting a t-test, it's crucial to ensure the data approximately follows a normal distribution. This is because the t-distribution, which is utilized for calculating the test, assumes normality. The t-test also takes into account the variability within the sample, which is measured by the sample standard deviation. This variability is reflected in the formula:\[t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\]where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. Understanding these components makes it easier to perform a t-test and interpret its results correctly.
null hypothesis
The null hypothesis is a foundational concept in hypothesis testing. It assumes no effect or no difference; in other words, any observed effect in an experiment is due to random chance. The purpose of hypothesis testing is to determine whether the data provides enough evidence to reject the null hypothesis.

In the original exercise, the null hypothesis \( H_0: \mu = 77 \) states that the average lifespan of Honolulu's residents is equal to 77 years. It's essentially a statement that there is no difference in lifespans from what has been previously reported. The alternative hypothesis, \( H_a: \mu < 77 \), opposes this by suggesting that the mean lifespan is less than 77 years.

Testing the null hypothesis requires evaluating whether observed data deviates significantly from the expectations of the null hypothesis. If the data shows a significant effect, we reject the null hypothesis in favor of the alternative hypothesis. However, if the results aren't significant, we fail to reject it, suggesting not enough evidence to support the alternative.
critical value
In hypothesis testing, a critical value is used to decide whether to reject the null hypothesis. It is determined based on the chosen significance level and the probability distribution of the test statistic. The critical value acts as a threshold. If the test statistic falls beyond this threshold, we reject the null hypothesis.

For the exercise provided, the critical value was derived from the t-distribution table, considering a one-tailed test at a 5% significance level with 19 degrees of freedom (since sample size - 1 equals 19). The critical value found was approximately -1.729.
  • A test statistic smaller than -1.729 would lead to rejecting the null hypothesis, indicating strong evidence that the mean lifespan is less than 77 years.
  • Conversely, if the test statistic is greater, as it was in this problem (-1.213), it doesn't reach the critical cutoff, and we fail to reject the null hypothesis.
Understanding critical values helps interpret the results of hypothesis tests and frames decisions around statistical evidence.
significance level
The significance level, often denoted by \( \alpha \), indicates the threshold at which we are willing to reject the null hypothesis. It reflects the probability of making a Type I error, which is rejecting a true null hypothesis. A common choice for a significance level is 5% (or 0.05), but it can be set at other levels like 1% or 10% depending on the context and requirements of the analysis.

In the exercise, a 5% significance level was chosen to test whether the lifespan in Honolulu is less than 77 years. This means the analysis would tolerate a 5% risk of incorrectly concluding that the average lifespan is shorter than 77 years when it is not.

Choosing an appropriate significance level is crucial as it influences the test’s sensitivity and specificity. A lower significance level makes it harder to reject the null hypothesis but decreases the chances of a Type I error. On the other hand, a higher significance level increases the likelihood of discovering a difference but at the cost of possibly making more Type I errors. Balance is key, as is considering the study’s consequences and context.

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Most popular questions from this chapter

Is the national crime rate really going down? Some sociologists say yes! They say that the reason for the decline in crime rates in the \(1980 \mathrm{~s}\) and \(1990 \mathrm{~s}\) is demographics. It seems that the population is aging, and older people commit fewer crimes. According to the FBI and the Justice Department, \(70 \%\) of all arrests are of males aged 15 to 34 years. (Source: True Odds, by J. Walsh, Merritt Publishing.) Suppose you are a sociologist in Rock Springs, Wyoming, and a random sample of police files showed that of 32 arrests last month, 24 were of males aged 15 to 34 years. Use a \(1 \%\) level of significance to test the claim that the population proportion of such arrests in Rock Springs is different from \(70 \%\).

When using a Student's \(t\) distribution for a paired differences test with \(n\) data pairs, what value do you use for the degrees of freedom?

Would you favor spending more federal tax money on the arts? This question was asked by a research group on behalf of The National Institute (Reference: Painting by Numbers, J. Wypijewski, University of California Press). Of a random sample of \(n_{1}=220\) women, \(r_{1}=\) 59 responded yes. Another random sample of \(n_{2}=175\) men showed that \(r_{2}=\) 56 responded yes. Does this information indicate a difference (either way) between the population proportion of women and the population proportion of men who favor spending more federal tax dollars on the arts? Use \(\alpha=0.05\).

In the following data pairs, A represents the cost of living index for housing and \(B\) represents the cost of living index for groceries. The data are paired by metropolitan areas in the United States. A random sample of 36 metropolitan areas gave the following information. (Reference: Statistical Abstract of the United States, 121 st edition.) \(\begin{array}{l|lllllllll} \hline A: & 132 & 109 & 128 & 122 & 100 & 96 & 100 & 131 & 97 \\ \hline B: & 125 & 118 & 139 & 104 & 103 & 107 & 109 & 117 & 105 \\ \hline A: & 120 & 115 & 98 & 111 & 93 & 97 & 111 & 110 & 92 \\ \hline B: & 110 & 109 & 105 & 109 & 104 & 102 & 100 & 106 & 103 \\ \hline \end{array}\) \(\begin{array}{l|rrrrrrrrr} \hline A: & 85 & 109 & 123 & 115 & 107 & 96 & 108 & 104 & 128 \\ \hline B: & 98 & 102 & 100 & 95 & 93 & 98 & 93 & 90 & 108 \\ \hline \\ \hline A: & 121 & 85 & 91 & 115 & 114 & 86 & 115 & 90 & 113 \\ \hline B: & 102 & 96 & 92 & 108 & 117 & 109 & 107 & 100 & 95 \\ \hline \end{array}\) i. Let \(d\) be the random variable \(d=A-B\). Use a calculator to verify that \(\bar{d} \approx 2.472\) and \(s_{d} \approx 12.124 .\) ii. Do the data indicate that the U.S. population mean cost of living index for housing is higher than that for groceries in these areas? Use \(\alpha=0.05\).

Let \(x\) be a random variable that represents red blood cell count \((\mathrm{RBC})\) in millions of cells per cubic millimeter of whole blood. Then \(x\) has a distribution that is approximately normal. For the population of healthy female adults, the mean of the \(x\) distribution is about \(4.8\) (based on information from Diagnostic Tests with Nursing Implications, Springhouse Corporation). Suppose that a female patient has taken six laboratory blood tests over the past several months and that the \(\mathrm{RBC}\) count data sent to the patient's doctor are \(\begin{array}{lllllll}4.9 & 4.2 & 4.5 & 4.1 & 4.4 & 4.3\end{array}\) i. Use a calculator with sample mean and sample standard deviation keys to verify that \(\bar{x}=4.40\) and \(s \approx 0.28\). ii. Do the given data indicate that the population mean \(\mathrm{RBC}\) count for this patient is lower than \(4.8\) ? Use \(\alpha=0.05\).

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