91Ó°ÊÓ

An article in Food Testing and Analysis ["Improving Reproducibility of Refractometry Measurements of Fruit Juices" (1999, Vol. 4, No. 4, pp. 13-17)] measured the sugar concentration (Brix) in clear apple juice. All readings were taken at \(20^{\circ} \mathrm{C}\) : $$ \begin{array}{lllll} 11.48 & 11.45 & 11.48 & 11.47 & 11.48 \\ 11.50 & 11.42 & 11.49 & 11.45 & 11.44 \\ 11.45 & 11.47 & 11.46 & 11.47 & 11.43 \\ 11.50 & 11.49 & 11.45 & 11.46 & 11.47 \end{array} $$ (a) Test the hypothesis \(H_{0}: \mu=11.5\) versus \(H_{1}: \mu \neq 11.5\) using \(\alpha=0.05 .\) Find the \(P\) -value (b) Compute the power of the test if the true mean is 11.4 . (c) What sample size would be required to detect a true mean sugar concentration of 11.45 if we wanted the power of the test to be at least \(0.9 ?\) (d) Explain how the question in part (a) could be answered by constructing a two-sided confidence interval on the mean sugar concentration. (e) Is there evidence to support the assumption that the sugar concentration is normally distributed?

Step by step solution

01

Calculate the Sample Mean and Standard Deviation

First, list all the measurements and determine the total number of observations, which is 20. Calculate the sample mean \(\bar{x}\) using the formula \(\bar{x} = \frac{\sum x_i}{n}\), where \(x_i\) are the measurements and \(n\) is the number of measurements. Next, calculate the sample standard deviation \(s\) using the formula \(s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}\). These calculations will provide the mean and variability which are key inputs for hypothesis testing.

02

Conduct the t-Test for Hypothesis (a)

Using the sample mean \(\bar{x}\), sample standard deviation \(s\), and sample size \(n\), perform a t-test. The test statistic is calculated using the formula \(t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}\), where \(\mu_0 = 11.5\) is the hypothesized population mean.Compare the calculated \(t\)-value to the critical \(t\)-value from the t-distribution at \(\alpha = 0.05\) and 19 degrees of freedom. If \(p\)-value < 0.05, reject the null hypothesis.

03

Calculate the P-value for Hypothesis (a)

The \(p\)-value is determined by finding the probability that a t-distributed random variable with 19 degrees of freedom is more extreme than the calculated \(t\)-value. Use statistical software or a t-table to find this probability. The \(p\)-value indicates the strength of evidence against the null hypothesis.

04

Compute the Power of the Test for (b)

The power of a test is the probability of correctly rejecting a false null hypothesis. Given a true mean \(11.4\), calculate the non-centrality parameter \(\delta\) using \(\delta = \frac{11.4 - 11.5}{s/\sqrt{n}}\). Next, use the software or a power table to find the power of the test at this non-centrality parameter, considering a significance level of \(0.05\).

05

Determine Required Sample Size for (c)

To determine the sample size required for a power of at least 0.9 to detect a difference of 0.05 (i.e., true mean of 11.45), use the formula for computing sample size: \( n = \left(\frac{(z_{\alpha/2} + z_{\beta}) \cdot s}{\mu_0 - \mu_1}\right)^2 \), where \(z_{\alpha/2}\) and \(z_{\beta}\) are the critical values corresponding to the desired confidence level and power. Solve for \(n\) using the known standard deviation \(s\).

06

Construct a Confidence Interval for (d)

A two-sided confidence interval for the mean can be constructed using: \[ \bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} \] where \(t_{\alpha/2}\) is the critical value from the t-distribution for 19 degrees of freedom. If \(11.5\) (null hypothesis mean) falls outside this interval, we have evidence to reject \(H_0\).

07

Assess Normality for (e)

Check if the data is normally distributed by using graphical methods like a Q-Q plot or statistical tests like the Shapiro-Wilk test for normality. If the plot shows data roughly fitting the line or the test has a \(p\)-value above \(0.05\), the normality assumption is considered reasonable.

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Key Concepts

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

t-test

Hypothesis testing often uses the t-test when the sample size is small, and the standard deviation of the population is unknown. The aim is to determine whether there is a significant difference between the sample mean and the hypothesized population mean.
We start by calculating the t-statistic using the formula: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] where \( \bar{x} \) is the sample mean, \( \mu_0 \) is the hypothesized mean, \( s \) is the sample standard deviation, and \( n \) is the sample size. Based on this statistic, we determine whether to reject the null hypothesis \( H_0 \) that the population mean is 11.5 g/mL.
If the calculated \( p \)-value is less than the significance level (\( \alpha = 0.05 \)), we reject the null hypothesis, suggesting that the true mean is significantly different from 11.5. This approach lets us evaluate claims about average measurements, like those in refractometry studies.

sample size determination

Determining the appropriate sample size is crucial for ensuring that a test or experiment has enough power to detect a true effect.
The power of a test, which should ideally be at least 0.9, is the probability of correctly rejecting a false null hypothesis. To find the necessary sample size \( n \) to achieve a desired power, use the following formula: \[ n = \left( \frac{(z_{\alpha/2} + z_{\beta}) \cdot s}{\mu_0 - \mu_1} \right)^2 \] where \( z_{\alpha/2} \) is the critical value for the confidence level, \( z_{\beta} \) is the critical value related to the power (usually from a 0.9 power level), \( s \) is the sample standard deviation, and \( \mu_0 - \mu_1 \) is the difference we want to detect.
This calculation ensures that the test is sensitive enough to reveal meaningful differences, supporting accurate conclusions in scientific research.

confidence interval

A confidence interval provides a range of values that is believed to contain the true population parameter with a certain level of confidence.
To construct a two-sided confidence interval for the mean sugar concentration, calculate: \[ \bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} \] where \( \bar{x} \) is the sample mean, \( t_{\alpha/2} \) is the critical value of the t-distribution at \( \alpha / 2 \) level of significance, and \( s \) is the sample standard deviation.
This interval provides an estimate of the range within which the true mean sugar concentration lies. If 11.5 (the hypothesized mean) is not within the confidence interval, it suggests that the null hypothesis might not be true. This method offers a visual and intuitive means of hypothesis testing, complementing the t-test.

normality assessment

Normality assessment is essential in statistical analysis, especially in the context of hypothesis testing with a t-test, which assumes normality.
To assess normality, graphical and statistical methods can be employed. A Q-Q plot, for instance, visually indicates how the sample data compares to a normal distribution. When data points fall along an approximate straight line, the normality assumption is met.
Additionally, the Shapiro-Wilk test provides a formal statistical test for normality. If the resulting \( p \)-value is greater than 0.05, the data do not deviate significantly from normality. However, violations do not always invalidate results as the t-test is robust to slight deviations with sufficiently large sample sizes. It's important to ensure data is reasonably normal to apply the t-test accurately in refractometry studies.