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solong [7]
2 years ago
5

A coin has heads on one side and tails on the other. The coin is tossed 12 times and lands

Mathematics
1 answer:
Fynjy0 [20]2 years ago
3 0

Answer:

D, the observed frequency of landing heads up gets closer to the expected frequency based on the probability of the coin landing heads up.

Step-by-step explanation:

As of now, the observed frequency states that the coin will land on heads 4/12 times. The expected frequency states that the coin will land on heads 50% of the time, or 6/12 times. As the number of trials increases, the observed frequency will get closer to the expected frequency.

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4.in a recent year, the population of california was about 2.52x10^7 people. its land area is about 4.05x10^5 km^2. what was the
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For the given hypothesis test, determine the probability of a Type II error or the power, as specified. A hypothesis test is to
erica [24]

Answer:

the probability of a Type II error if in fact the mean waiting time u, is 9.8 minutes is 0.1251

Option A) is the correct answer.

Step-by-step explanation:

Given the data in the question;

we know that a type 11 error occur when a null hypothesis is false and we fail to reject it.

as in it in the question;

obtained mean is 9.8 which is obviously not equal to 8.3

But still we fail to reject the null hypothesis says mean is 8.3

Hence we have to find the probability of type 11 error

given that; it is right tailed and o.5, it corresponds to 1.645

so

z is equal to 1.645

z = (x-μ)/\frac{S}{\sqrt{n} }

where our standard deviation s = 3.8

sample size n = 50

mean μ = 8.3

we substitute

1.645 = (x - 8.3)/\frac{3.8}{\sqrt{50} }

1.645 = (x - 8.3) / 0.5374

0.884023 = x - 8.3

x = 0.884023 + 8.3

x = 9.18402

so, by general rule we will fail to reject the null hypothesis when we will get the z value less than 1.645

As we reject the null hypothesis for right tailed test when the obtained test statistics is greater than the critical value

so, we will fail to reject the null hypothesis as long as we get the sample mean less than 9.18402

Now, for mean 9.8 and standard deviation 3.8 and sample size 50

Z =  (9.18402 - 9.8)/\frac{3.8}{\sqrt{50} }

Z = -0.61598 / 0.5374

Z = - 1.1462 ≈ - 1.15

from the z-score table;

P(z<-1.15) = 0.1251

Therefore, the probability of a Type II error if in fact the mean waiting time u, is 9.8 minutes is 0.1251

Option A) is the correct answer.

8 0
2 years ago
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