The mean Income per peron In the Unted States Is 40,500, and the distribution of Incomes follows a normal distribution. A randornsample of 18 resldents of Wilmington, Deaware, hod a rmean of $51,500 with a standard deviation of $10 200. At the 0.050 level ofsigntficance, E that enough cvidence to conclude that reoldents of Wilmington, Delawsrc, have more tncome than the national226euDADState the null hypothests and the alternate hypothests.H1 P?b. State the decislon rule for 0.050 elgntficance level. (Round your answer to 3 dectmal pleces.)ప్పయు 3c.Compute the value of the test stot'stic (Round your answer to 2 decimel places.)of Witmlngtan. Dabware. bave more Income than the nat onsl average atate thot r455u29no vBncuc c.ouaspthe 0.0s0 s gnfcance

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Asked Nov 24, 2019
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The mean Income per peron In the Unted States Is 40,500, and the distribution of Incomes follows a normal distribution. A randorn
sample of 18 resldents of Wilmington, Deaware, hod a rmean of $51,500 with a standard deviation of $10 200. At the 0.050 level of
signtficance, E that enough cvidence to conclude that reoldents of Wilmington, Delawsrc, have more tncome than the national
226euDAD
State the null hypothests and the alternate hypothests.
H1 P?
b. State the decislon rule for 0.050 elgntficance level. (Round your answer to 3 dectmal pleces.)
ప్పయు 3
c.Compute the value of the test stot'stic (Round your answer to 2 decimel places.)
of Witmlngtan. Dabware. bave more Income than the nat onsl average at
ate thot r
455u29no vBncuc c.ouasp
the 0.0s0 s gnfcance
help_outline

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The mean Income per peron In the Unted States Is 40,500, and the distribution of Incomes follows a normal distribution. A randorn sample of 18 resldents of Wilmington, Deaware, hod a rmean of $51,500 with a standard deviation of $10 200. At the 0.050 level of signtficance, E that enough cvidence to conclude that reoldents of Wilmington, Delawsrc, have more tncome than the national 226euDAD State the null hypothests and the alternate hypothests. H1 P? b. State the decislon rule for 0.050 elgntficance level. (Round your answer to 3 dectmal pleces.) ప్పయు 3 c.Compute the value of the test stot'stic (Round your answer to 2 decimel places.) of Witmlngtan. Dabware. bave more Income than the nat onsl average at ate thot r 455u29no vBncuc c.ouasp the 0.0s0 s gnfcance

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Expert Answer

Step 1

a.

The null and alternative hypothesis is obtained below:

Hypotheses:

Null hypothesis:

H0: µ ≤ 40,500

Alternative hypothesis:

H1: µ > 40,500

Since the alternative hypothesis states µ > 40,500, this test is a right-tailed test.

b.

The sample size (n) is 18

Degrees of freedom

Df = n – 1

     = 18 –1

     = 17

Step-by-step procedure to obtain the critical value using MINITAB software:

  • Choose Graph > Probability Distribution Plot choose View Probability > OK.
  • From Distribution, choose ‘t’ distribution.
  • In Degrees of freedom, enter 17.
  • Click the Shaded Area
  • Choose P Value and Right Tail for the region of the curve to shade.
  • Enter the data value as 0.05.
  • Click OK.

Output obtained using MINITAB software is given below:

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Distribution Plot T, df 17 0.4 0.3 0.2 0.1 0.05 0.0 0 1.740 Density ox

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Step 2

Here, the sample mean, x-bar is 51,500.

Population mean, µ is 40,500.

Sample standard deviation, s = 10,200

Sample size, n is 18.

The test statistic for t-test is calculated as 4.58 from the calculations given below.

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51,500-40,500 10,200 V18 11,000 2,404.1631 = 4.58

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Step 3

d.

 Computation of P-value:

The right tailed P-value for the t-test is 0.000 which can be obtained using the excel formula, &ld...

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T.DIST.RT(4.58,17) D 0.000133

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