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noname [10]
3 years ago
13

A student group has $6,000 budgeted for a field trip. The cost of transportation for the trip is $2,800. To stay within the budg

et, all other costs C must be no more than what amount (in $)?
Mathematics
1 answer:
docker41 [41]3 years ago
8 0

Answer:c=3200

Step-by-step explanation:

2800+c<=6000

2800-2800+c<=6000-2800

c=3200

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The test statistic for the appropriate test is \frac{0.15-0.275}{\sqrt{0.2125(1-0.2125)[\frac{1}{40}+\frac{1}{40}]}}.

Step-by-step explanation:

The experiment conducted here is to determine whether there is a difference in proportion of starfish over 8 inches in length in the two ocean areas, i.e. north and south.

The hypothesis to test this can be defined as follows:

<em>H₀</em>: There is no difference in proportion of starfish over 8 inches in length in the two ocean areas, i.e. <em>p</em>₁ = <em>p</em>₂.

<em>Hₐ</em>: There is a difference in proportion of starfish over 8 inches in length in the two ocean areas, i.e. <em>p</em>₁ ≠ <em>p</em>₂.

The two-proportion <em>z</em>-test would be used to perform the test.

A sample of <em>n</em> = 40 starfishes are selected from both the ocean areas.

It provided that of the 40 starfish from the north, 6 were found to be over 8 inches in length and of  the 40 starfish from the south, 11 were found to be over 8 inches in length.

Compute the sample proportion of starfish from north that were over 8 inches in length as follows:

\hat p_{n}=\frac{6}{40}=0.15

Compute the sample proportion of starfish from south that were over 8 inches in length as follows:

\hat p_{s}=\frac{11}{40}=0.275

The test statistic is:

z=\frac{\hat p_{n}-\hat p_{s}}{\sqrt{P(1-P)[\frac{1}{n_{n}}+\frac{1}{n_{s}}]}}

Compute the combined proportion <em>P</em> as follows:

P=\frac{X_{n}+X_{s}}{n_{n}+n_{s}}=\frac{6+11}{40+40}=0.2125

Compute the test statistic value as follows:

z=\frac{\hat p_{n}-\hat p_{s}}{\sqrt{P(1-P)[\frac{1}{n_{n}}+\frac{1}{n_{s}}]}}

  =\frac{0.15-0.275}{\sqrt{0.2125(1-0.2125)[\frac{1}{40}+\frac{1}{40}]}}

Thus, the test statistic for the appropriate test is \frac{0.15-0.275}{\sqrt{0.2125(1-0.2125)[\frac{1}{40}+\frac{1}{40}]}}.

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