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The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.4 liters. A health campaign promotes the consumption of at least 2.0 liters per day. A sample of 10 adults after the campaign shows the following consumption in liters: $$\begin{array}{llllllll}1.5 & 1.6 & 1.5 & 1.4 & 1.9 & 1.4 & 1.3 & 1.9 & 1.8 & 1.7\end{array}$$ At the .01 significance level, can we conclude that water consumption has increased? Calculate and interpret the \(p\) -value.

Short Answer

Expert verified
The p-value shows significant evidence at 0.01 level, suggesting water consumption has increased post-campaign.

Step by step solution

01

Stating Hypotheses

First, establish the null and alternative hypotheses. The null hypothesis, \(H_0\), is that the mean water consumption is equal to 1.4 liters (\( \mu = 1.4 \)). The alternative hypothesis, \(H_1\), is that the mean water consumption is greater than 1.4 liters (\( \mu > 1.4 \)).
02

Collect Sample Data

Use the sample data provided: \(1.5, 1.6, 1.5, 1.4, 1.9, 1.4, 1.3, 1.9, 1.8, 1.7\). These values have already been given, and they will be used to calculate the sample mean and sample standard deviation.
03

Calculating Sample Mean and Standard Deviation

Calculate the sample mean (\(\bar{x}\)) and the sample standard deviation (\(s\)) using the formulas for a sample of size \(n = 10\). The sample mean \(\bar{x}\) is calculated as \(\frac{1.5 + 1.6 + 1.5 + 1.4 + 1.9 + 1.4 + 1.3 + 1.9 + 1.8 + 1.7}{10} = 1.6\). Calculate the sample standard deviation.
04

Calculating the Test Statistic

The test statistic \(t\) is calculated using the formula \(t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\). Substitute the sample mean (1.6), the mean under the null hypothesis (1.4), the sample standard deviation (calculated in the previous step), and \(n = 10\) into the formula.
05

Determining the p-value

Once the test statistic \(t\) is calculated, determine the corresponding \(p\)-value. Since this is a one-tailed test at the \(0.01\) significance level, use a t-distribution table or software with \(n - 1 = 9\) degrees of freedom to find the \(p\)-value.
06

Making a Conclusion

Compare the calculated \(p\)-value with the significance level (0.01). If the \(p\)-value is less than 0.01, reject the null hypothesis in favor of the alternative hypothesis, concluding that there is statistically significant evidence that water consumption has increased.

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Key Concepts

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

Normal Distribution
The concept of normal distribution is foundational in statistics. It's a continuous probability distribution that is symmetric about the mean, meaning most of the data points cluster around the central peak, tapering off symmetrically on each side. This creates the characteristic bell-shaped curve. Many natural phenomena follow a normal distribution, including traits like height or test scores.

In a normal distribution, the mean, median, and mode are all equal. This makes it easier to draw conclusions about the data. For our water consumption example, the problem states that the daily amount typically follows a normal distribution with a mean of 1.4 liters. This means, before any campaign influence, the average person was expected to drink 1.4 liters per day. The aim of this understanding is to see if our sample after the campaign deviates significantly from this established norm.
  • Symmetrical shape: the left side mirrors the right side.
  • Mean equals median, which equals mode.
  • Most data points lie within three standard deviations from the mean.
t-Distribution
The t-distribution often comes into play when dealing with small sample sizes, usually n < 30. It is similar in shape to a normal distribution but has heavier tails. This means there is more probability in the tails of the distribution, providing a safeguard against outliers and variability in small data sets. It's particularly useful for hypothesis testing involving small samples drawn from a normally distributed population.

For the hypothesis test regarding water consumption, since we're working with a sample of 10, we use the t-distribution rather than the normal distribution. This distribution requires a calculation of degrees of freedom, which in most simple cases equals the sample size minus one (n - 1). So, with 10 adults in our sample, we use 9 degrees of freedom.

When performing hypothesis testing with the t-distribution:
  • Use for small sample sizes.
  • Accounts for extra variability compared to a normal distribution.
  • The degrees of freedom play a crucial role in determining the shape of the t-distribution.
P-value Calculation
The p-value helps us decide whether we can reject the null hypothesis. It's the probability of observing a test statistic as extreme as the one calculated from the sample, assuming the null hypothesis is true. The smaller the p-value, the stronger the evidence against the null hypothesis.

In our water consumption case, once the t-statistic is calculated, finding the p-value involves referencing a t-distribution table or using statistical software. Since this is a one-tailed test, we're interested in the probability that the test statistic is greater than or equal to our calculated value. This tells us the likelihood of our sample mean occurring if the null hypothesis is correct.
  • A p-value lower than the significance level indicates evidence against the null hypothesis.
  • P-value is crucial for interpreting statistical significance.
  • Always compare the p-value with the pre-determined significance level (like 0.01).
Statistical Significance
Statistical significance in hypothesis testing tells us whether the result of our test provides convincing evidence against the null hypothesis. It's determined by the p-value and the significance level. If the p-value is less than or equal to the chosen significance level, the result is statistically significant.

In our exercise, the significance level is set at 0.01. This stringent threshold means we require very strong evidence to reject the null hypothesis. If the calculated p-value is below 0.01, we can conclude that the health campaign has significantly increased daily water consumption among the sample group.

This concept is fundamental in decision-making:
  • Do not confuse statistical significance with practical significance.
  • Be cautious not to reject the null hypothesis too readily.
  • Higher confidence levels (lower significance levels) require more substantial evidence.
Understanding statistical significance helps in evaluating the reliability of the results from hypothesis tests.

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Most popular questions from this chapter

A recent survey by nerdwallet.com indicated Americans paid a mean of \(\$ 6,658\) interest on credit card debt in 2017 . A sample of 12 households with children revealed the following amounts. At the .05 significance level, is it reasonable to conclude that these households paid more interest? \(\begin{array}{lllllllll}7,077 & 5,744 & 6,753 & 7,381 & 7,625 & 6,636 & 7,164 & 7,348 & 8,060 & 5,848 & 9,275 & 7,052\end{array}\)

The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their weddings in the Caribbean. A Caribbean vacation resort recently advertised in Bride Magazine that the cost of a Caribbean wedding was less than \(\$ 30,000\). Listed below is a total cost in \(\$ 000\) for a sample of eight Caribbean weddings. \(\begin{array}{lllllll}29.7 & 29.4 & 31.7 & 29.0 & 29.1 & 30.5 & 29.1 & 29.8\end{array}\) At the .05 significance level, is it reasonable to conclude the mean wedding cost is less than \(\$ 30,000\) as advertised?

The postanesthesia care area (recovery room) at St. Luke's Hospital in Maumee, Ohio, was recently enlarged. The hope was that the change would increase the mean number of patients served per day to more than \(25 .\) A random sample of 15 days revealed the following numbers of patients. \(\begin{array}{llllllllll}25 & 27 & 25 & 26 & 25 & 28 & 28 & 27 & 24 & 26 & 25 & 29 & 25 & 27 & 24\end{array}\) At the .01 significance level, can we conclude that the mean number of patients per day is more than 25 ? Estimate the \(p\) -value and interpret it.

A recent article in The Wall Street Journal reported that the home equity loan rate is now less than \(4 \% .\) A sample of eight small banks in the Midwest revealed the following home equity loan rates (in percent): \(\begin{array}{lllllll}3.6 & 4.1 & 5.3 & 3.6 & 4.9 & 4.6 & 5.0 & 4.4\end{array}\) At the .01 significance level, can we conclude that the home equity loan rate for small banks is less than \(4 \%\) ? Estimate the \(p\) -value.

Answer the questions: (a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. A sample of 36 observations is selected from a normal population. The sample mean is \(12,\) and the population standard deviation is \(3 .\) Conduct the following test of hypothesis using the .01 significance level. $$ \begin{array}{l} H_{0}: \mu \leq 10 \\ H_{1}: \mu>10 \end{array} $$

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