Question

a sample mean, sample size, and sample standard deviation are provided below. use the one - mean t - test to perform the required hypothesis test at the 10% significance level.
\\( \bar { x } = 26, s = 9, n = 32, h _ { 0 } : \mu = 21, h _ { a } : \mu > 21 \\)
click here to view a partial table of values of \\( t _ { \alpha } \\).
the test statistic is \\( t = \square \\).
(round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for the one - mean t - statistic

The formula for the one - mean t - statistic is \(t=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu_0\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step2: Substitute the given values into the formula

Given \(\bar{x} = 26\), \(\mu_0=21\), \(s = 9\), and \(n = 32\).
First, calculate \(s/\sqrt{n}\): \(\frac{s}{\sqrt{n}}=\frac{9}{\sqrt{32}}\approx\frac{9}{5.65685}\approx1.591\)
Then, calculate \(t\): \(t=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}=\frac{26 - 21}{9/\sqrt{32}}=\frac{5}{9/\sqrt{32}}\)
\(t=\frac{5\sqrt{32}}{9}=\frac{5\times5.65685}{9}\approx\frac{28.28425}{9}\approx3.14\)

Answer:

\(3.14\)