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Fiesta28 [93]
3 years ago
6

A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought i

n to the time repairs are completed. A random sample of 12 repair records showed the following repair times (in days): 5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.
H0: \mu \leq 5 days versus H1: \mu > 5 days. At \alpha = .05, choose the right option.
a) Reject H0 if tcalc < 1.7960
b) Reject H0 if tcalc >1.7960
Mathematics
1 answer:
garri49 [273]3 years ago
6 0

Answer:

The degrees of freedom first given by:  

df=n-1=12-1=11  

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

t_{\alpha}= 1.796

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

Step-by-step explanation:

Information given

5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.

System of hypothesis

We want to test if the true mean is higher than 5, the system of hypothesis are :  

Null hypothesis:\mu \leq 5  

Alternative hypothesis:\mu > 5  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

The degrees of freedom first given by:  

df=n-1=12-1=11  

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

t_{\alpha}= 1.796

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

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Suppose a basketball player has made 231 out of 361 free throws. If the player makes the next 2 free throws, I will pay you $31.
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Answer:

The expected value of the proposition is of -0.29.

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either the player will make it, or he will miss it. The probability of a player making a free throw is independent of any other free throw, which means that the binomial probability distribution is used to solve this question.

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.

Suppose a basketball player has made 231 out of 361 free throws.

This means that p = \frac{231}{361} = 0.6399

Probability of the player making the next 2 free throws:

This is P(X = 2) when n = 2. So

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

P(X = 2) = C_{2,2}.(0.6399)^{2}.(0.3601)^{0} = 0.4095

Find the expected value of the proposition:

0.4095 probability of you paying $31(losing $31), which is when the player makes the next 2 free throws.

1 - 0.4059 = 0.5905 probability of you being paid $21(earning $21), which is when the player does not make the next 2 free throws. So

E = -31*0.4095 + 21*0.5905 = -0.29

The expected value of the proposition is of -0.29.

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