Question

Measurements of water intake, obtained from a sample of 17 rats that had been injected with a sodium chloride solution, produced a mean and standard deviation of 31.0 and 6.2 cubic centimeters \(\left(\mathrm{cm}^{3}\right),\) respectively. Given that the average water intake for noninjected rats observed over a comparable period of time is \(22.0 \mathrm{~cm}^{3},\) do the data indicate that injected rats drink more water than noninjected rats? Test at the \(5 \%\) level of significance. Find a \(90 \%\) confidence interval for the mean water intake for injected rats.

Short answer

Additionally, find the 90% confidence interval for the mean water intake for injected rats. Answer: Based on the comparison of the test statistic and the critical value, we can conclude that injected rats drink more water than noninjected rats at a 5% level of significance. The 90% confidence interval for the mean water intake for injected rats is approximately 31 ± (1.746 * 6.2 / √17).

Step-by-step solution

State the null and alternative hypotheses

First, we need to state the null hypothesis (H0) and the alternative hypothesis (H1). H0: The mean water intake for injected rats is equal to the mean water intake for noninjected rats (µ = 22 cm³) H1: The mean water intake for injected rats is greater than the mean water intake for noninjected rats (µ > 22 cm³)

Calculate the test statistic

Next, we will calculate the test statistic using the following formula: t = (sample mean - population mean) / (standard deviation / √n) where n is the sample size. In this case: t = (31 - 22) / (6.2 / √17)

Find the critical value

For a one-sample t-test, the critical value is determined by the degrees of freedom (n - 1) and the desired level of significance (5% in this case). For this exercise, the degrees of freedom is: df = 17 - 1 = 16 Using a t-table or calculator, we find the critical value for a one-tailed test with df = 16 and α = 0.05 to be approximately 1.746.

Compare test statistic and critical value

Now, we will compare the test statistic and the critical value to make a decision about the null hypothesis. If t > critical value, we reject the null hypothesis (H0) and conclude that injected rats drink more water than noninjected rats.

Calculate the 90% confidence interval

To calculate the 90% confidence interval for the mean water intake for injected rats, we use the following formula: Confidence interval = sample mean ± (t_critical * standard deviation / √n) where t_critical is the critical value from the t-distribution corresponding to the desired confidence level (90% in this case) and the degrees of freedom (df = 16). For a 90% confidence level, the t_critical value is approximately 1.746 as before. Confidence interval = 31 ± (1.746 * 6.2 / √17)

Key concepts

Confidence Interval

A confidence interval gives us a range of values in which we can be fairly sure the true mean of a population lies. It is a vital part of statistical inference. The width of the interval depends on the variability of the data and the size of the sample.

To calculate a confidence interval, we need to know:

• The sample mean, which is the average of the measurements from your sample.
• The standard deviation, which measures the spread of the sample data.
• The sample size, which is the number of observations in your sample.
• The t-value, determined by the chosen confidence level and the number of degrees of freedom.

In our example, we calculate a 90% confidence interval for the average water intake of injected rats. We use the formula:\[\text{Confidence interval} = \text{sample mean} \pm (t_{critical} \times \text{standard deviation}/\sqrt{n})\]This interval indicates that we are 90% confident that the true mean water intake for injected rats falls within this range.

One-Sample t-Test

A one-sample t-test is a statistical test used to determine whether there is a significant difference between the mean of a single sample and a known or hypothesized population mean.

The purpose of this test is straightforward: assess whether injected rats have different water intakes from non-injected ones. The mathematical magic happens in a few steps:

• We begin by stating the null hypothesis: the sample mean equals the population mean. Here, we hypothesize that the water intake of injected rats is equal to that of non-injected rats.
• The alternative hypothesis is what you think is true. In our case, that injected rats drink more.
• We use the test statistic formula: \( t = \frac{\text{sample mean} - \text{population mean}}{\text{standard deviation}/\sqrt{n}} \).
• Finally, we compare the calculated t-value with the critical t-value, which helps us decide if we reject the null hypothesis.

In our scenario, after calculating the t-value and comparing it with the critical value from a t-table, we make an informed decision regarding the water intake.

Significance Level

The significance level, often denoted as \( \alpha \), is a threshold set by researchers to decide whether a result is statistically significant. A common choice for \( \alpha \) is 0.05, or 5%, which means there is a 5% risk of concluding that a difference exists when there is none.

The choice of \( \alpha \) affects our decision in hypothesis testing:

• If our test statistic exceeds the critical value at the \( \alpha \) level, we reject the null hypothesis.
• This means we are reasonably sure that any observed difference is not due to random chance.
• Conversely, if the test statistic is less than the critical value, we do not reject the null hypothesis, indicating that the data does not provide strong evidence against it.

In our given problem, with \( \alpha = 0.05 \), we're stating that we need at least 95% confidence in the results before asserting that injected rats drink more than non-injected ones. The significance level thus helps balance the possibility of making errors in hypothesis testing.