07.06.2023

State the hypothesis and claim: The expense of moving the storage yard for the consolidated Package Delivery Service is justified only if it can be shown that the daily average travel distance is less than 214 mi. In 12 trial runs the average distance was 198 mi with a standard deviation of 42 mi At a = .01 test the claim that the average daily travel distance is less than 214 mi.

the claim is . H0: ___ H1:

. 0

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24.06.2023, solved by verified expert
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Null Hypothesis: H₀ : μ < 214 miles

Alternative Hypothesis :H₁: μ> 214 miles

The calculated value |t| = 1.319 < 3.4966 at 0.01 level of significance

Null hypothesis is accepted

The average daily travel distance is less than 214 miles

Step-by-step explanation:

Step(i):-

Given that the size of the sample 'n' = 12

The mean of the Population = 214 miles

The mean of the sample   = 198 miles

The standard deviation of the sample = 42 miles

Level of significance = 0.01

Step(ii):-

Null Hypothesis: H₀ : μ < 214 miles

Alternative Hypothesis :H₁: μ> 214 miles

Test statistic

             State the hypothesis and claim: The expense of, №17886920, 07.06.2023 14:44

             State the hypothesis and claim: The expense of, №17886920, 07.06.2023 14:44

          |t| = |- 1.319| = 1.319

Degrees of freedom   = n-1 = 12 -1 =11

critical value  State the hypothesis and claim: The expense of, №17886920, 07.06.2023 14:44

The calculated value |t| = 1.319 < 3.4966 at 0.01 level of significance

Null hypothesis is accepted

The average daily travel distance is less than 214 miles

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Mathematics
Step-by-step answer
P Answered by PhD

Null Hypothesis: H₀ : μ < 214 miles

Alternative Hypothesis :H₁: μ> 214 miles

The calculated value |t| = 1.319 < 3.4966 at 0.01 level of significance

Null hypothesis is accepted

The average daily travel distance is less than 214 miles

Step-by-step explanation:

Step(i):-

Given that the size of the sample 'n' = 12

The mean of the Population = 214 miles

The mean of the sample   = 198 miles

The standard deviation of the sample = 42 miles

Level of significance = 0.01

Step(ii):-

Null Hypothesis: H₀ : μ < 214 miles

Alternative Hypothesis :H₁: μ> 214 miles

Test statistic

             t = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }

             t = \frac{198 -214}{\frac{42}{\sqrt{12} } }

          |t| = |- 1.319| = 1.319

Degrees of freedom   = n-1 = 12 -1 =11

critical value  t_{(0.005 , 11)} =   3.4966

The calculated value |t| = 1.319 < 3.4966 at 0.01 level of significance

Null hypothesis is accepted

The average daily travel distance is less than 214 miles

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