A supermarket is redesigning it’s checkout lanes. Design A has a sample size of 50, sample mean of 4.1 minutes, and sample standard deviation of 2.2 minutes. Design B has a sample size of 50, sample mean of 3.5 minutes, and sample standard deviation of 1.5 minutes. At the 0.05 level of significance, determine if their is evidence that the checkout times of the two systems differ.

The calculated value t = 1.57736 < 1.9845 at 5 % level of significance

Null hypothesis is accepted at 5 % level of significance

There is no significance difference between Design A and Design B

Step-by-step explanation:

Given  sample size of design A

                                            n₁ = 50

sample mean of design A x⁻₁ = 4.1 minutes

Sample standard deviation S₁ = 2.2 minutes

Given  sample size of design B

                                            n₂ = 50

sample mean of design A x⁻₂ = 3.5 minutes

Sample standard deviation S₂ = 1.5 minutes

Null Hypothesis : H₀ : There is no significance difference between Design A and Design B

Alternative Hypothesis : H₁:There is  significance difference between Design A and Design B

Level of significance ∝ = 0.05

Test statistic

where

On calculation , we get

S² = 3.6173

Test statistic

On calculation , we get

   t = 1.57736

Degrees of freedom

ν = n₁ + n₂ -2 = 50 +50 -2 =98

t₀.₀₂₅ ,₉₈ = 1.9845

The calculated value t = 1.57736 < 1.9845 at 5 % level of significance

null hypothesis is accepted

step-by-step explanation:

30% of $420 is 0.30($420) = $126 and $420 + $126 = $546.

or, 130% of $420 is 1.3($420) = $546.

notice that this is not the $600 original list price with which we started.

why doesn't this work?

the method does not work because 30% of $420 is not the same amount as the 30% of $600. it may seem obvious when stated this way, but it is a very common error made on problems of this type. we cannot just use the same percent and get back to the original amount because we are taking that percent of a different value.

to work problems of this type correctly, it is generally best to write an equation for the situation and then solve.

example: since (100% – 30%)(original price) = (sale price)

70% of the   (original price) is $420.

0.7x   = 420

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x = 600

the original cost was $600.

we may also solve this problem using the two-step method. working the problem in two steps requires us to use more algebra skills. we have to know that x = 1x (identity property of multiplicationtext annotation indicator) and that 1x – 0.3x = 0.7x (subtracting like terms). then we solve the problem as follows:

(original price) – 30%(original price) = sale price

x – 0.3x = 420

0.7x = $420

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x = 600

the original cost was $600.  

note that with both of the above methods we do obtain the correct original amount.

example: after a 5% pay raise, hermione is earning $22,680 per year. what was she earning before the pay raise?

105% of (original salary) = current salary

1.05x = 22,680

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x = 21,600

hermione earned $21,600 per year before the pay raise.

this is an example so maybe you will get how to at least do it!

about 0.75 hours

step-by-step explanation: