Respuesta :
Answer:
The sample does not contradicts the manufacturer's claim.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 200 mg
Sample mean, [tex]\bar{x}[/tex] = 211.5 mg
Sample size, n = 20
Alpha, α = 0.05
Sample standard deviation, s = 18.5 mg
a) First, we design the null and the alternate hypothesis for a two tailed test
[tex]H_{0}: \mu = 200\\H_A: \mu \neq 200[/tex]
We use Two-tailed t test to perform this hypothesis.
b) Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex] Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{211.5 - 200}{\frac{18.5}{\sqrt{20}} } = 2.7799[/tex]
c) Now,
[tex]t_{critical} \text{ at 0.05 level of significance, 19 degree of freedom } = \pm 2.0930[/tex]
Since, the calculated t-statistic does not lie in the range of the acceptance region(-2.0930,2.0930), we reject the null hypothesis.
Thus, the sample does not contradicts the manufacturer's claim.
d) P-value = 0.011934
Since the p-value is less than the significance level, we reject the null hypothesis and accept the alternate hypothesis.
Yes, both approach the critical value and the p-value approach gave the same results.
