A convenience sample differs from a voluntary sample in that a convenience sample is structured based on accessibility to the researcher, and a voluntary sample is based on participant interest. convenience samples survey each participant once, and voluntary samples survey each participant numerous times. convenience sampling is a method of random sampling, and a voluntary sample is not. convenience sampling is not a probability-based method, and voluntary sampling is.

\( 1 \leftarrow \begin{array}{l}\text { Researchers conducted a study on using a chief executive officer's facial structure to predict a firm's financial } \\ \text { performance. The facial width-to-height ratio (WHR) for each in a sample of } 52 \text { CEOs at publicly traded firms was } \\ \text { determined. The sample resulted in } \bar{x}=1.55 \text { and } s=0.36 \text {. An analyst wants to predict the financial performance of a } \\ \text { firm based on the value of the true mean facial WHR of CEOs. The analyst wants to use the value of } \mu=2.1 \text {. Do yo } \\ \text { recommend he use this value? Conduct a test of hypothesis for } \mu \text { to help you answer the question. Specify all the } \\ \text { elements of the test, including } H_{0}, H_{a} \text {, test statistic, } p \text {-value, and your conclusion. Test at } \alpha=0.01 \text {. } \\ \text { What is the value of the test statistic? } \\ z=-11.02 \text { (Round to two decimal places as needed.) } \\ \text { Find the p-value. } \\ \text { p-value }=\square \text { (Round to three decimal places as needed.) }\end{array} \).

A random sample of 69 produced the summary statistics \( \bar{x}=0.326 \) and \( \mathrm{s}^{2}=0.064 \). Complete parts a and \( b \). a. Test the null hypothesis that \( \mu=0.37 \) against the alternative hypothesis that \( \mu<0.37 \), using \( \alpha=0.10 \). What is the value of the test statistic? \( z=-1.44 \) (Round to two decimal places as needed.) Find the \( p \)-value. p-value \( =\square \) (Round to three decimal places as needed.)

A random sample of 69 produced the summary statistics \( \bar{x}=0.326 \) and \( s^{2}=0.064 \). Complete parts a and \( b \). State and interpret the results of the test. A. Reject \( H_{0} \). There is insufficient evidence to indicate that the true population mean is less than 0.37 at \( \alpha=0.10 \). Bo not reject \( H_{0} \). There is sufficient evidence to indicate that the true population mean is less than 0.37 at \( \alpha=0.10 \). C. Reject \( H_{0} \). There is sufficient evidence to indicate that the true population mean is less than 0.37 at \( \alpha=0.10 \). Do not reject \( H_{0} \). Thefe is insufficient evidence to indicate that the true population mean is less than 0.37 at \( \alpha=0.10 \).

A variable \( x \) is normally distributed with mean 21 and standard deviation 8 . Round your answers to the nearest hundredth as needed. a) Determine the \( z \)-score for \( x=27 \). \[ z=0.75 \] b) Determine the \( z \)-score for \( x=16 \). \[ z= \]

In a normal distribution, a data value located 1.2 standard deviations below the mean has Standard Score: \( \mathrm{z}=\square \)