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miv72 [106K]
3 years ago
11

You are the manager of a restaurant that delivers pizza to college dormitory rooms. You have just changed your delivery process

in an effort to reduce the mean time between the order and completion of delivery from the current 21 minutes. A sample of 36 orders using the new delivery process yields a sample mean of 22.4 minutes and a sample standard deviation of 6 minutes.
A.) Please compute the test statistic. Please round to four decimal places.
B.) Please determine the p-value. Please round to four decimal places.
C.) Using a 0.05 level of significance, please provide the critical value for the appropriate test. Please round to four decimal places.
D.) Using a 0.05 level of significance, is there evidence that the population mean delivery time has been reduced below the previous population mean value of 25 minutes?
Mathematics
1 answer:
lozanna [386]3 years ago
4 0

Answer:

1.4000 ;

0.0807 ;

1.69 ;

Fail to reject the Null

Step-by-step explanation:

Given :

μ = 21 minutes

x = 22.4 minutes

Sample size, n = 36

Standard deviation, s = 6

H0: μ ≥ 21

H1: μ < 21

Test statistic : (x - μ) ÷ s/sqrt(n)

Test statistic : (22.4 - 21) ÷ 6/sqrt(36)

Test statistic : 1.4 ÷ 1

Test statistic = 1.4000

Pvalue : Using the Test statistic value at α = 0.05 ; One - tail test = 0.0807

The t critical value using a critical value calculator ; DF = (n - 1) = 36 - 1 = 35 ; α = 0.05 = 1.69

Decision region :

Reject H0 if

|Test statistic| > |critical value|

1.4 < 1.69

Hence, we fail to reject H0

There isn't enough evidence they delivery time has been reduced below previous mean

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Answer:

The price per wind chime that will maximize revenue = $ 315

Step-by-step explanation:

Given - A gift shop sells 160 wind chimes per month at $150 each. the owners estimate that for each $15 increase in price, they will sell 5 fewer wind chimes per month.

To find - Find the price per wind chime that will maximize revenue.

Proof -

Given that,

Total Wind chimes selling = 160

Price of each Wind chime = $150

Now,

Given that, for each $15 increase in price, they will sell 5 fewer wind chimes per month.

So,

Let the price = 150 + 15x

So,

Number of wind Chimes sold per month = 160 - 5x

So,

Total Revenue, R = (150 + 15x)(160 - 5x)

                             = 24000 - 750x + 2400x - 75x²

                             = 24000 + 1650x - 75x²

⇒R(x) = 24000 + 1650x - 75x²

Differentiate R with respect to x , we get

R'(x) = 1650 - 150x

Now,

For Maximize Revenue, Put R'(x) = 0

⇒1650 - 150x = 0

⇒150x = 1650

⇒x = 1650/150

⇒x = 11

∴ we get

Price per Wind chime = $ 150 + 15(11)

                                    = $ 150 + 165

                                    = $ 315

So,

The price per wind chime that will maximize revenue = $ 315

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