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The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes. The p-value for this test statistics is: a. 0.05 b. 0.0228 c. 0.0456 d. 0.025 brainly

Sagot :

Cricetus

Answer:

The appropriate answer is option b (0.0228).

Step-by-step explanation:

The given values are:

Random sample,

[tex]n=100[/tex]

Average length of time,

[tex]\bar{x}=3.1 \ minutes[/tex]

Waiting time,

[tex]\mu=3 \ minutes[/tex]

Standard deviation,

[tex]\sigma=0.5[/tex]

Now,

The test statistics will be:

⇒  [tex]z=\frac{(\bar{x}-\mu)}{\frac{\sigma}{\sqrt{n} } }[/tex]

On substituting the values, we get

⇒     [tex]=\frac{(3.1-3)}{\frac{0.5}{\sqrt{100} } }[/tex]

⇒     [tex]=\frac{0.1}{0.05}[/tex]

⇒     [tex]=2[/tex]

hence,

⇒  [tex]P(z >2 ) = 1 - P(z <2 )[/tex]

By using the table,

⇒                 [tex]=1-0.9772[/tex]

⇒                 [tex]=0.0228[/tex]

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