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A sample of 16 ATM transactions shows a mean transaction time of 67 seconds with a sample standard deviation of 12 seconds. State the hypotheses to prove that the mean transaction time exceeds 60 seconds. Assume that times are normally distributed.a. Determine your hypotheses.b. Compute the test statistic. What’s the rejection rule?d. At the α =.05 level of significance, your Critical Value ise. What conclusion can be drawn from this test at a 0.05 significance level?

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Answer:

We conclude that  the mean transaction time exceeds 60 seconds.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 60 seconds

Sample mean, [tex]\bar{x}[/tex] = 67 seconds

Sample size, n = 16

Alpha, α = 0.05

Sample standard deviation, s = 12 seconds

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 60\text{ seconds}\\H_A: \mu > 60\text{ seconds}[/tex]

We use One-tailed(right) t test to perform this hypothesis.

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{67 - 60}{\frac{12}{\sqrt{16}} } = 2.34[/tex]

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 15 degree of freedom } = 1.753[/tex]

Since,                  

[tex]t_{stat} > t_{critical}[/tex]

We fail to accept the null hypothesis and reject it. Thus, we conclude that  the mean transaction time exceeds 60 seconds.

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