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Effectus [21]
3 years ago
12

A researcher wants to estimate the mean number of hours per day that elementary aged school children spend watching TV in order

to direct subsequently planned research in reducing time spent watching TV among young children. She collects a sample of 424 children and calculates a 95% confidence interval for number of hours per day spent watching TV: Confidence interval - (1.93, 4.25).
a. (3pts) Interpret this confidence interval in a sentence below.


b. (5pts) Would a two-sided hypothesis test, performed at significance level 0.05, for testing whether mean number of hours per day spent watching TV is equal to 2.5. reject the null hypothesis? How about a test for mean number of hours per day being equal to 1.2? (You need to have two answers for this question. The first is for p = 2.5, and the second is for u = 1.2).
Mathematics
1 answer:
Mashutka [201]3 years ago
3 0

Answer:

a)  The confidence interval (1.93, 4.25) means that there is 95% confidence that the population mean for the number of hours per day spent watching TV is within 1.93 and 4.25.

b) H0: μ=2.5. The null hypothesis failed to be rejected.

H0: μ=1.2. The null hypothesis is rejected.

Step-by-step explanation:

a)  The confidence interval (1.93, 4.25) means that there is 95% confidence that the population mean for the number of hours per day spent watching TV is within 1.93 and 4.25.

b) If a two-sided hypthesis test is performed at a significance level of 0.05 with a null hypothesis H0: μ=2.5, the result would be that the null hypothesis failed to be rejected.

This is because we have the information of the 95% confidence level, that has the same level of significance and is also two-sided, includes the value 2.5. The confidence interval is, in this case, equivalent to the acceptance region for the hypothesis test, so the null hypothesis failed to be rejected.

If the null hypothesis mean would have fallen outside of the limits of the confidecence interval, the null hypothesis would have been rejected.

This is the case of H0: μ=1.2, which lies outside the confidence interval (in the rejection region). Then, the null hypothesis is rejected.

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a) 0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

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Step-by-step explanation:

For itens a and b, we use conditional probability.

For item c, we use the binomial distribution along with the conditional probability.

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

a) If it has an emergency locator, what is the probability that it will not be discovered?

Event A: Has an emergency locator

Event B: Not located.

Probability of having an emergency locator:

66% of 72%(Are discovered).

20% of 100 - 72 = 28%(not discovered). So

P(A) = 0.66*0.72 + 0.2*0.28 = 0.5312

Probability of having an emergency locator and not being discovered:

20% of 28%. So

P(A cap B) = 0.2*0.28 = 0.056

Probability:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.056}{0.5312} = 0.105

0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) If it does not have an emergency locator, what is the probability that it will be discovered?

Probability of not having an emergency locator:

0.5312 of having. So

P(A) = 1 - 0.5312 = 0.4688

Probability of not having an emergency locator and being discovered:

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P(A \cap B) = 0.34*0.72 = 0.2448

Probability:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2448}{0.4688} = 0.522

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P(A \cap B) = 0.66*0.72 = 0.4752

Probability of being discovered, given that it has the emergency recorder:

p = P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4752}{0.5312} = 0.8946

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P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 7) = C_{10,7}.(0.8946)^{7}.(0.1054)^{3} = 0.064

0.064 = 6.4% probability that 7 of them are discovered.

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