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A simple random sample of 100 concert tickets was drawn from a normal population. The mean and standard deviation of the sample were $120 and $25, respectively. Test the hypothesis H0: µ = 125 vs. H1: µ ≠ 125 at the 1% significance level. Rejection region: | t | > t0.05,99 = 1.66 Test statistic: t = -2.0 p-value one tail= 0.024 p-value two tail= 0.048

Respuesta :

Answer:

We accept the null hypothesis and the population mean is $120.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 100

Sample mean, [tex]\bar{x}[/tex] = $120

Alpha, α = 0.01

Sample standard deviation, s = $25

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 125\\H_A: \mu \neq 125[/tex]

We use two-tailed t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = -2.0[/tex]              

p-value one tail= 0.024

p-value two tail= 0.048

Conclusion:

Since the p-value for two tailed test is greater than the significance level, we fail to reject the null hypothesis and accept it.

Thus, the population mean is $120.