91Ó°ÊÓ

For Exercises 7 through \(23,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. Water Consumption The Old Farmer's Almanac stated that the average consumption of water per person per day was 123 gallons. To test the hypothesis that this figure may no longer be true, a researcher randomly selected 16 people and found that they used on average 119 gallons per day and \(s=5.3 .\) At \(\alpha=0.05,\) is there enough evidence to say that the Old Farmer's Almanac figure might no longer be correct? Use the \(P\) -value method.

Step by step solution

01

State the Hypotheses

We want to test if the average water consumption per person per day is different from 123 gallons, as stated in the Old Farmer's Almanac.Null Hypothesis (H_0): \( \mu = 123 \) Alternative Hypothesis (H_1): \( \mu eq 123 \)Here, the claim is that the current population mean is different from 123 gallons, which is what we are testing against the null hypothesis.

02

Find the Critical Value(s)

Since the sample size is small (N = 16) and the population standard deviation is unknown, we use the t-distribution. The degrees of freedom (df) is calculated as N - 1 = 15.For a two-tailed test with \alpha = 0.05, the critical t-values are found using a t-table or calculator.Critical t-values: \( _{\alpha/2, 15} = \pm 2.131 \).

03

Calculate the Test Statistic

The test statistic is calculated using the formula:\[t = \frac{\bar{x} - \mu}{s/\sqrt{N}},\]where \(\bar{x} = 119\), \(\mu = 123\), \(s = 5.3\), and \(N = 16\).\[t = \frac{119 - 123}{5.3/\sqrt{16}} = \frac{-4}{1.325} \approx -3.02\]

04

Make the Decision

Compare the calculated t-value with the critical t-values.Since \(-3.02 < -2.131\), the calculated t-value falls in the rejection region of the t-distribution.Therefore, we reject the null hypothesis.

05

Summarize the Results

At a significance level of \(\alpha = 0.05\), there is enough evidence to reject the null hypothesis. This suggests that the average water consumption per person per day could be different from the 123 gallons claimed by the Old Farmer's Almanac.

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Key Concepts

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

Null Hypothesis

The null hypothesis is a foundational concept in hypothesis testing. It is an initial assumption we make about the population parameter. In our exercise, the null hypothesis, denoted as \( H_0 \), assumes that the average water consumption per person per day is 123 gallons, as historically claimed by the Old Farmer's Almanac.

This hypothesis acts as a baseline, suggesting there is no change or effect. In hypothesis testing, we often look to gather evidence to see if the null hypothesis can be rejected. When setting up a test:

• We determine what we are testing. Here, if the actual water usage differs from 123 gallons.
• We also define the alternative hypothesis, \( H_1 \), which suggests that the average is not equal to 123 gallons.

The null hypothesis provides a point of comparison, and it’s essential because it helps us understand the initial claim or assumption about a population parameter.

Critical Value

The critical value is a threshold that determines the boundary between the region where we would reject the null hypothesis and where we would not. In many hypothesis tests, including this one where we've applied the t-distribution due to the small sample size, critical values are derived based on the chosen significance level \( \alpha \).

When calculating critical values, we:

• Use the t-distribution if the population standard deviation is unknown and the sample size is small (less than 30).
• Determine degrees of freedom \(df\), typically as the sample size minus one.
• Use statistical tables or software to find critical values based on the degrees of freedom and significance level.

In this scenario, with \(df = 15\) and \(\alpha = 0.05\), the critical t-values are \(\pm 2.131\). Our calculated test statistic will be compared against these values to decide whether to reject the null hypothesis.

Test Statistic

The test statistic is calculated to assess the validity of the null hypothesis. It quantifies how far our sample statistic is from the hypothesized population parameter under the null hypothesis, in units of standard error.

To find the test statistic, use:

• The formula \[ t = \frac{\bar{x} - \mu}{s/\sqrt{N}} \] where \( \bar{x} \) is the sample mean (average consumption), \( \mu \) is the hypothesized population mean (123 gallons), \( s \) is the sample standard deviation, and \( N \) is the sample size.
• Insert our values: \( \bar{x} = 119 \), \( \mu = 123 \), \( s = 5.3 \), \( N = 16\).

When the test statistic is calculated, we get approximately \(-3.02\). This value helps us understand if our observed data significantly departs from the null hypothesis expectations.

P-value Method

The p-value method is a common approach in hypothesis testing, allowing us to determine statistical significance. A p-value represents the probability that the observed data occurred by random chance, given that the null hypothesis is true.

Here's how the p-value method works:

• A small p-value (typically \( \leq 0.05 \)) indicates strong evidence against the null hypothesis. This suggests that the observed effect is significant.
• To compute the p-value, we assess where our test statistic lies within the t-distribution. The lower p-value corresponds to a test statistic falling in the extreme tails of the distribution.
• If the p-value \( < \alpha \), we reject the null hypothesis.

In our problem, since the calculated test statistic was \(-3.02\) and it falls in the rejection region, it implies a small p-value, justifying the rejection of the null hypothesis. This methodology provides valuable insights into whether the historical claim about water consumption holds or requires revision.