Question

test the claim about the population mean \\( \mu \\) at the level of significance \\( \alpha \\). assume the population is normally distributed

claim: \\( \mu < 5015 \\); \\( \alpha = 0.01 \\) \quad sample statistics: \\( \bar{x} = 5117 \\), \\( s = 5725 \\), \\( n = 54 \\)

...

(type integers or decimals. do not round.)

find the standardized test statistic \\( t \\).

\\( t = 0.13 \\)

(round to two decimal places as needed.)

find the p - value.

\\( p = \square \\)

(round to three decimal places as needed.)

Explanation:

Step1: Identify the test type

This is a one - tailed t - test (since the claim is $\mu<5015$) with $n = 54$, so the degrees of freedom $df=n - 1=54 - 1 = 53$. The test statistic $t = 0.13$.

Step2: Use t - distribution to find P - value

For a one - tailed t - test with $t = 0.13$ and $df = 53$, we can use a t - distribution table or a statistical software. Using a t - distribution calculator (or software like Excel's T.DIST function), the formula for the P - value of a left - tailed test is $P=P(T

Using the t - distribution, for $t = 0.13$ and $df = 53$, we calculate the P - value. Using a calculator or software (e.g., in R, pt(0.13,53)), we get the P - value.

Calculating: The t - distribution with $df = 53$ and $t = 0.13$. The cumulative distribution function for the t - distribution gives the probability that $T5015$, so the P - value will be greater than 0.5. Wait, no, the test statistic is calculated as $t=\frac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}}$, where $\mu_0 = 5015$, $\bar{x}=5117$, $s = 5725$, $n = 54$. Let's recalculate the test statistic to check:

$t=\frac{5117 - 5015}{\frac{5725}{\sqrt{54}}}=\frac{102}{\frac{5725}{7.3485}}\approx\frac{102}{779.07}\approx0.131$, which is approximately 0.13.

For a left - tailed test ( $H_0:\mu\geq5015$, $H_a:\mu<5015$), the P - value is $P(T<0.13|df = 53)$. Since the t - distribution is symmetric around 0, $P(T<0.13)=0.5 + P(0

Answer:

The P - value is approximately $\boxed{0.551}$ (rounded to three decimal places).