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daser333 [38]
3 years ago
14

A small diner has one employee and a counter with seating for 8 customers. The diner does not package food for takeout. Customer

s arrive at the diner at the rate of 20 per hour (Poisson distributed). Service times are exponentially distributed and average 24 per hour. Customers that arrive when all seats are taken do not enter the diner. What is the average time a customer spends in the diner?
Select one:
A. 3 minutes
B. 5.975 minutes
C. 6.44 minutes
D. 8.94 minutes
Mathematics
1 answer:
Fed [463]3 years ago
8 0

Answer:

Step-by-step explanation:

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An education researcher claims that 58​% of college students work​ year-round. In a random sample of 400 college​ students, 232
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Answer:

The proportion of college students who work​ year-round is 58%.

Step-by-step explanation:

The claim made by the education researcher is that 58​% of college students work​ year-round.

A random sample of 400 college​ students, 232 say they work​ year-round.

To test the researcher's claim use a one-proportion <em>z</em>-test.

The hypothesis can be defined as follows:

<em>H</em>₀: The proportion of college students who work​ year-round is 58%, i.e. <em>p</em> = 0.58.

<em>Hₐ</em>: The proportion of college students who work​ year-round is 58%, i.e. <em>p</em> ≠ 0.58. C

Compute the sample proportion as follows:

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Compute the test statistic value as follows:

 z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.58-0.58}{\sqrt{\frac{0.58(1-0.58)}{400}}}=0

The test statistic value is 0.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

Compute the p-value for the two-tailed test as follows:

 p-value=2\times P(z

*Use a z-table for the probability.

The p-value of the test is 1.

The p-value of the test is very large when compared to the significance level.

The null hypothesis will not be rejected.

Thus, it can be concluded that the proportion of college students who work​ year-round is 58%.

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