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With individual lines at its various windows, a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. To test this claim, the post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes.With a significance level of 5 percent, test the claim that a single line causes lower variation among waiting times (shorter waiting times) for customers. Degrees of freedom test statistic pvalue accept or reject

Respuesta :

Answer:

We reject H₀ we found that at 5% significance level the single file causes shorther waiting times

Step-by-step explanation:

We will develop a chi-square test since it is a standard deviation test.

Conditions for chi-square test

1.-Population follows a normal distribution

2.-We have a random sample

Therefore

Test Hypothesis

Null hypothesis                           H₀            σ   =  7,2

Alternative hypothesis               Hₐ            σ   <  7,2

So it is a one tailed-test to the left

Sample size      n  = 25

Then df  =  25 - 1      df  = 24

Significance level  α = 5%    α = 0,05

With   df = 24  and   α = 0,05  we find

Χ₀ = 36,415  

Computing  X(s)

X(s)  = ( n - 1 ) * σ² / (7,2)²

X(s)  =  24* (3,5)² / 51,84

X(s)  =  294/51,84

X(s)  =  5,67

Comparing     Χ₀     and      Χ(s)

X(s) <   Χ₀

So X(s) is in the rejection region and we reject H₀

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