Question

a random sample of 78 eighth grade students scores on a national mathematics assessment test has a mean score of 277. this test result prompts a state school administrator to declare that the mean score for the states eighth graders on this exam is more than 270. assume that the population standard deviation is 36. at \\( \alpha = 0.08 \\), is there enough evidence to support the administrators claim? complete parts (a) through (e).
(a) write the claim mathematically and identify \\( h_0 \\) and \\( h_a \\). choose the correct answer below.
a. \\( h_0: \mu \leq 270 \\) (claim)
\\( h_a: \mu>270 \\)
b. \\( h_0: \mu \geq 270 \\) (claim)
\\( h_a: \mu<270 \\)
c. \\( h_0: \mu \leq 270 \\)
\\( h_a: \mu>270 \\) (claim)
d. \\( h_0: \mu = 270 \\)
\\( h_a: \mu>270 \\) (claim)
e. \\( h_0: \mu = 270 \\) (claim)
\\( h_a: \mu>270 \\)
f. \\( h_0: \mu<270 \\)
\\( h_a: \mu \geq 270 \\) (claim)
(b) find the standardized test statistic z.
\\( z = 1.72 \\) (round to two decimal places as needed.)
(c) find the p - value.
\\( p - value = \\) (round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for P - value in a right - tailed z - test

For a right - tailed z - test, \(P - value=P(Z > z)\), where \(Z\) is a standard normal random variable and \(z\) is the standardized test statistic.

Step2: Use the standard normal distribution table or a calculator

We know that \(z = 1.72\). Using a standard normal distribution table or a calculator with the normalcdf function (\(normalcdf(lower,upper,0,1)\)), for a right - tailed test with \(z = 1.72\), we calculate \(P(Z>1.72)\).
Since the total area under the standard normal curve is \(1\), and \(P(Z\leq z)\) can be found from the standard normal table. \(P(Z\leq1.72)=0.9573\) (from the standard normal table). Then \(P(Z > 1.72)=1 - P(Z\leq1.72)\).

$$P(Z>1.72)=1 - 0.9573=0.043$$

Answer:

\(0.043\)