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BARSIC [14]
3 years ago
14

An automobile assembly line operation has a scheduled mean completion time, , of minutes. The standard deviation of completion t

imes is minutes. It is claimed that, under new management, the mean completion time has decreased. To test this claim, a random sample of completion times under new management was taken. The sample had a mean of minutes. Can we support, at the level of significance, the claim that the mean completion time has decreased under new management
Mathematics
1 answer:
fomenos3 years ago
8 0

The question is incomplete. The complete question is :

An automobile assembly line operation has a scheduled mean completion time, μ, of 15.5 minutes. The standard deviation of completion times is 1.7 minutes. It is claimed that, under new management, the mean completion time has decreased. To test this claim, a random sample of 90 completion times under new management was taken. The sample had a mean of 15.4 minutes. Can we support, at the 0.1 level of significance, the claim that the mean completion time has decreased under new management?

Solution :

The given data :

n = 90

μ = 15.5

σ = 1.7

$\overline x$ = 15.4

So, the null hypothesis is : $H_0: \mu = 15.5$ is been tested against

Alternate hypothesis : $H_1 : \mu < 15.5 $   (one-tailed test)

Since, the sample size is sufficiently large and the sample deviation is known, we use the Z-test.

Under this test, the test statistic is given as :

$Z=\frac{\overline x - \mu}{\sigma/\sqrt{n}} \sim N(0,1)$

Under H_0, we have

$Z_0=\frac{\overline x - \mu}{\sigma/\sqrt{n}} $

$Z_0=\frac{15.4-15.5}{1.7/\sqrt{90}} $

$Z_0=-0.56$

The critical value for the test is $Z_{\alpha} = Z_{0.1} = -1.28$

We observe that (-0.56 > -1.28), and so we fail to reject the null hypothesis.

No, there is no evidence to support the claim that the mean completion time has decreased.

Thus, we conclude that the mean completion time is 15.5 minutes.

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Answer:

3 groups

Step-by-step explanation:

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12 = 1, 2, 3, 4, 6, 12

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The highest common factor of 12 and 9 is 3

Therefore, the choir teacher can only have 3 groups

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Each of the 3 groups Will have 4 sopranos and 3 altos each

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kow [346]
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150/6 = 25
184 + 52 = 236
4 0
3 years ago
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In the diagram shown, FJ bisects AD BL~=LC. The length of CD is 2 more than the length of LC, and the length of AD is
LekaFEV [45]

Answer:

The Length of AC is  17 units.

Step-by-step explanation:

Given:

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By Addition Property we have

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