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kupik [55]
3 years ago
15

A person in the audience of a talk show has a 15% chance of winning free movie tickets. What is the chance that a person in the

audience may not win free movie tickets?
85%
70%
35%
30%
Mathematics
2 answers:
Sergeu [11.5K]3 years ago
7 0
-- The person in the audience will either win tickets or not win tickets.

-- The probability of (either win tickets or not win tickets) is 100%.

-- (Probability of win tickets) plus (probability of not win tickets) = 100%.

-- If (probability of win tickets) is 15%, then the (probability of not win tickets)
is the other 85%.
LenKa [72]3 years ago
5 0

Answer:A. 85% hope it helps ig idk

Step-by-step explanation:

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Answer:

t=\frac{44-42}{\frac{1.9}{\sqrt{40}}}=6.657    

p_v =P(t_{(39)}>6.657)=3.17x10^{-8}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 42 at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=44 represent the sample mean

s=1.9 represent the sample standard deviation

n=40 sample size  

\mu_o =42 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

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We need to conduct a hypothesis in order to check if the mean is higher than 42, the system of hypothesis would be:  

Null hypothesis:\mu \leq 42  

Alternative hypothesis:\mu > 42  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{44-42}{\frac{1.9}{\sqrt{40}}}=6.657    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=40-1=39  

Since is a one side test the p value would be:  

p_v =P(t_{(39)}>6.657)=3.17x10^{-8}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 42 at 1% of signficance.  

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