A distributor of personal computers has five locations in a city. In the year's first quarter, the sales in units were: Location Observed Sales (units) North side 70 Pleasant Township 75 Southwick 70 I-90 50 Venice Avenue 35 Total 300 For a goodness-of-fit test that sales were the same for all locations, what is the critical value at the 0.01 level of risk

It is observed  that the value of test statistics (19.168) is greater than the critical value (13.277), thus the rejected hypothesis, H₀ at α = 0.01.

There is enough evidence to conclude or deduce that sales were the same for all locations

Explanation:

Solution

Given that:

Object: Test whether the sales were the same for all locations by applying 1% significance level.

The Null Hypothesis H₀ : Sales were the same for all locations

E₁ =  70 +75 +70+ 50 + 35/5 = 60

The Alternative Hypothesis Hₐ : Sales were not the same for all locations

Now,

The decision rule:

the Level of Significance be α = 0.01

Degrees of freedom is df= Number of categories -1

=5-1 = 4

Note: Kindly find an attached copy of part of  the work solution of this given question

x=3

step-by-step explanation:

this is really simple i saw that there was only one term (5x) and thought if that was 15 what would i have to multiply everything else to get it there? well if 5x became 15 then x would have to be 3 so plug the 3 in the x for the other equations and you get 15 for the other ones. so there you go. : )

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