Question

test the claim about the population mean μ at the level of significance α. assume the population is normally distributed. claim: μ < 5215; α = 0.01. sample statistics: x̄ = 5317, s = 5949, n = 53. find the standardized test statistic t. t = 0.12 (rounded to two decimal places as needed). find the p - value. p = (rounded to three decimal places as needed).

Explanation:

Step1: Recall the formula for the t - statistic

The formula for the t - statistic in a one - sample t - test is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\). Here, \(\bar{x} = 5317\), \(\mu = 5215\), \(s = 5849\), and \(n = 53\).

$$t=\frac{5317 - 5215}{5849/\sqrt{53}}$$

$$t=\frac{102}{5849/7.2801}$$

$$t=\frac{102}{803.48}$$

\(t\approx0.13\) (rounded to two decimal places)

Step2: Find the degrees of freedom

The degrees of freedom for a one - sample t - test is \(df=n - 1\). So, \(df=53-1 = 52\)

Step3: Use technology or a t - table to find the P - value

Since this is a left - tailed test (\(H_a:\mu<5215\)), we use the t - distribution with \(df = 52\). Using a t - calculator or software (e.g., in R: pt(0.13,52)), the P - value is the probability \(P(T<t)\) where \(T\) follows a t - distribution with \(df = 52\).
\(P\approx0.552\) (rounded to three decimal places)

Answer:

\(P = 0.552\)