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 = 5049, n = 53

find the standardized test statistic t.
t = 0.12
(round to two decimal places as needed.)

find the p - value.
p = 0.552
(rounded to three decimal places as needed.)

decide whether to reject or fail to reject the null hypothesis. choose the correct answer below.
fail to reject h₀. there is enough evidence at the is level of significance to support the claim.

Explanation:

Step1: State the null and alternative hypotheses

The claim is \(\mu = 5215\). So, \(H_0:\mu = 5215\) and \(H_1:\mu
eq5215\) (two - tailed test).

Step2: Recall the formula for the t - test statistic

The formula for the t - test statistic is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\). Given \(\bar{x} = 5317\), \(\mu = 5215\), \(s = 5949\), \(n = 53\).

$$t=\frac{5317 - 5215}{5949/\sqrt{53}}$$

$$t=\frac{102}{5949/7.2801}$$

$$t=\frac{102}{817.16}$$

\(t\approx0.125\approx0.13\) (rounded to two decimal places)

Step3: Find the degrees of freedom

Degrees of freedom \(df=n - 1=53-1 = 52\)

Step4: Find the P - value

Using a t - distribution table or a calculator (e.g., in R: \(2*(1 - pt(0.13,52))\)), the P - value is approximately \(0.897\) (using a more accurate calculation:
The cumulative distribution function of the t - distribution \(F(t;df)\) gives \(P(T<0.13;52)\approx0.5515\). For a two - tailed test \(P = 2*(1 - 0.5515)=0.897\))

Step5: Make a decision

Since \(\alpha=0.01\) and \(P = 0.897>0.01\), we fail to reject \(H_0\).

Answer:

We fail to reject \(H_0\). There is not enough evidence at the \(0.01\) level of significance to reject the claim.