Question

test the claim about the population mean \\( \mu \\) at the level of significance \\( \alpha \\). assume the population is normally distributed.\
claim: \\( \mu \geq 1473 \\), \\( \alpha = 0.03 \\), \\( \sigma = 26 \\). sample statistics: \\( \bar{x} = 1465 \\), \\( n = 25 \\)\
\\( h_0: \mu \geq 1465 \\)\
\\( h_a: \mu < 1465 \\)\
calculate the standardized test statistic.\
the standardized test statistic is \\( -1.54 \\) (round to two decimal places as needed.)\
determine the p - value.\
\\( p = 0.0618 \\) (round to three decimal places as needed.)\
determine the outcome and conclusion of the test.\
fail to reject \\( h_0 \\). at the 3% significance level, there is not enough evidence to reject the claim.\
an example get more help \\( \boldsymbol{\cdot} \\)\
clear all final check

Explanation:

Step1: Identify Test Type

This is a left - tailed z - test (since population standard deviation \(\sigma\) is known, and we are testing a claim about the population mean \(\mu\)). The null hypothesis \(H_0:\mu\geq1473\) and the alternative hypothesis \(H_a:\mu < 1473\) (wait, there is a typo in the original problem, the sample mean \(\bar{x} = 1465\), and the hypothesized mean in \(H_0\) should be related. Let's assume the correct null hypothesis is \(H_0:\mu\geq1473\) and \(H_a:\mu < 1473\)). The formula for the z - test statistic is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\)

Step2: Calculate the z - statistic

Given \(\bar{x}=1465\), \(\mu = 1473\), \(\sigma = 26\), \(n = 25\)
First, calculate the standard error \(SE=\frac{\sigma}{\sqrt{n}}=\frac{26}{\sqrt{25}}=\frac{26}{5} = 5.2\)
Then, the z - statistic \(z=\frac{1465 - 1473}{5.2}=\frac{- 8}{5.2}\approx - 1.54\) (matches the given test statistic)

Step3: Calculate the P - value

For a left - tailed test, the P - value is \(P(Z < z)\), where \(z=-1.54\). Using the standard normal distribution table or calculator, \(P(Z < - 1.54)=0.0618\) (matches the given P - value)

Step4: Compare P - value with \(\alpha\)

Given \(\alpha = 0.03\). The P - value \(0.0618>0.03\). So we fail to reject the null hypothesis \(H_0\). This means at the 3% significance level, there is not enough evidence to reject the claim (since the claim is in the null hypothesis \(H_0:\mu\geq1473\))

Answer:

The standardized test statistic is \(-1.54\), the P - value is \(0.0618\), we fail to reject \(H_0\), and at the 3% significance level, there is not enough evidence to reject the claim.