Answer:
Step-by-step explanation:
Given that:
The population mean [tex]\mu[/tex] = 30
The sample size n = 200
The sample-mean [tex]\overline x[/tex] = 27.5
The sample standard deviation s = 18.9
The level of significance ∝ = 0.05
The null hypothesis and the alternative hypothesis can be written as:
[tex]H_o:\mu \ge 30 \\ \\ H_1: \mu < 30[/tex]
Since the alternative hypothesis indicates < sign, then this test is left-tailed
The t-test statistics is:
[tex]t =\dfrac{\overline x - \mu }{\dfrac{s}{\sqrt{n}}}[/tex]
[tex]t =\dfrac{27.5 -30 }{\dfrac{18.9}{\sqrt{200}}}[/tex]
[tex]t =\dfrac{-2.5}{\dfrac{18.9}{14.142}}[/tex]
t = -1.871
The p-value = P(t< -1.871)
From the standard normal t tables
The p-value = 0.0314
Thus, since the p-value is less than the level of significance ∝ = 0.05, then we reject the null hypothesis.
