A local retailer claims that the mean waiting time is less than 8 minutes. A random sample of 20 waiting times has a mean of 6.3 minutes with a standard deviation of 2.1 minutes. At α = 0.01, test the retailer's claim. Assume the distribution is normally distributed. Round the test statistic to the nearest thousandth.

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Answer:

The claim is not true

Step-by-step explanation:

We are given that A local retailer claims that the mean waiting time is less than 8 minutes.

[tex]H_0:\mu=8[/tex]

[tex]H_a:\mu<8[/tex]

A random sample of 20 waiting times has a mean of 6.3 minutes with a standard deviation of 2.1 minutes.

[tex]\bar{x}=6.3[/tex]

s = 2.1

n = 20

Since n <30 and population standard deviation is unknown

So,we will use t test

So,[tex]t=\frac{x-\mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t=\frac{6.3-8}{\frac{2.1}{\sqrt{20}}}[/tex]

t=-3.62

α = 0.01

Degree of freedom = df=n-1=20-1=19

[tex]t_{df,\frac{\alpha}{2}}=t_{19,\frac{0.01}{2}}=2.861[/tex]

t calculated < t critical

So, we failed to reject null hypothesis

Hence the claim is not true

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