9 Consider the following hypotheses H0 5600 The population

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Consider the following hypotheses:

H₀: ? = 5,600

The population is normally distributed with a population standard deviation of 620. Compute the value of the test statistic and the resulting p-value for each of the following sample results. For each sample, determine if you can \"reject/do not reject\" the null hypothesis at the 10% significance level. Use Table 1.(Negative values should be indicated by a minus sign. Round intermediate calculations to 4 decimal places. Round \"test statistic\" values to 2 decimal places and \"p-value\" to 4 decimal places.)

Solution

Answer to the question)

We got

standard deviaton s = 620

M = 5600

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The formula of test statistic is :

z = (x bar - M ) / (s/sqrt(n))

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Part a)
we got x bar = 5660

THus Ho: M = 5600

Ha: M > 5600

[right tailed test]

n = 145

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On plugging the values we get

z = (5660-5600) / (620/sqrt(145))

z = 1.67

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P value = 1 - 0.8790 = 0.1210 [this can be obtained from the z table]

Since the P value 0.1210 > alpha 0.10 , we FAIL to reject the null

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Part b)

Again since the value of x bar is larger than M , we consider it to be a right tailed test

z = (5660 -5600) / (620/sqrt(325))

z = 1.74

P value =1- 0.9599 = 0.0401

P value 0.0401 < 0.10 , we REJECT the null hypothesis

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part c)

In this case the vlaue of xbar 5310 < 5600 , hence we consider it to be a left tailed test

z = (5310 -5600) / (620/sqrt(33))

z = -2.69

Pvalue = 0.0036 .....[this value we can directly get from the z table , because the z table tells the left area]

Result: P value 0.0036 < alpha 0.10 , We REJECT the null

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Part d)

it is again left tailed test

z = (5350-5600) / (620/sqrt(33))

z = -2.32

P value = 0.1017

Result : Since the P value 0.1017 > alpha 0.10 , We FAIL to reject the null

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Part c)