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QveST [7]
3 years ago
10

A method currently used by doctors to screen patients for a certain type of cancer fails to detect cancer in 15% of the patients

who actually have the disease. A new method has been developed that researchers hope will be able to detect cancer more accurately. A random sample of 80 patients known to this type of cancer is screened using the new method and the method failed to detect the cancer in 8 patients. At the 5% level of significance, can the researchers conclude that the new method is better than the one currently in use? (Can they conclude that the new method fails to detect cancer in less than 15% of the patients who actually have the disease?)
Mathematics
1 answer:
Novosadov [1.4K]3 years ago
8 0

Answer:

A technique presently used through docs to display screen ladies for breast most cancers fails to observe most cancers in 15% of the girls who without a doubt have the disease. A random pattern of eighty girls acknowledged to have breast most cancers are to be screened the use of a new technique that researchers suppose will be capable to realize most cancers greater accurately. At the 0.05 stage of significance, the researchers will be capable to conclude that the new approach is higher than the present day one if the fantastic.

Answer::: greater than 1.645

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What is the value of x if 8:15 ::16x
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Step-by-step explanation:

8:15::16:x

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A shirt for $40, jeans for $50 and a watch for $35 are purchased at the mall. If 7% sales tax is charged, what will the total co
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35 + 40 + 50= 125$

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Which is a better estimate for the length of a hockey stick?
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I would say either inches of feet.

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A box contains 5 tickets numbered 1,2,3,4, and 5. Two tickets are drawn at random from the box. Find the chance that the numbers
adelina 88 [10]

Answer:

(a) The probability that the numbers on the two tickets differ by two or more if the draws are made with replacement is 0.48.

(b) The probability that the numbers on the two tickets differ by two or more if the draws are made without replacement is 0.60.

Step-by-step explanation:

The tickets are drawn such that the difference between the two numbers is at least 2.

That is, 1st number - 2nd number ≥ 2.

The sample space such that this condition is satisfied is:

S = {(1, 3), (1, 4), (1, 5), (2, 4), (2, 5) and (3, 5)} = 6 possible pairs.

But there is also case where the numbers can be drawn in reverse order, i.e we can draw (3, 1) instead of (1, 3).

This makes the total number of possible pairs as, 6 × 2 = 12 pairs.

(a) <u>With Replacement</u>

In case the tickets are selected with replacement, then the probability of selecting the 1st ticket is same as for the 2nd ticket is:

P(Difference\geq 2)=P(1st\ ticket)\times P(2nd\ ticket)\times No.\ of\ Possible\ pairs

                               =\frac{1}{5}\times \frac{1}{5}\times12\\ =0.48

Thus, the probability that the numbers on the two tickets differ by two or more if the draws are made with replacement is 0.48.

(b) <u>Without Replacement:</u>

In case the tickets are selected without replacement, then the probability of selecting the 1st ticket is same as for the 2nd ticket is:

P(Difference\geq 2)=P(1st\ ticket)\times P(2nd\ ticket)\times No.\ of\ Possible\ pairs

                               =\frac{1}{5}\times \frac{1}{4}\times12\\ =0.60

Thus, the probability that the numbers on the two tickets differ by two or more if the draws are made without replacement is 0.60.

7 0
3 years ago
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