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2 years ago

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You and a friend have a hula hoop competition. you win 3 out of every 7 trials. your friend wins 5 more trials than you. how man

y trials are in the competition?

1 answer:

Ivenika [448]2 years ago

3 0

Answer:

Total number of trials is 35.

Step-by-step explanation:

Given : Me and my friend have a hula hoop competition. I win 3 out of every 7 trials. My friend wins 5 more trials than me.

To find : How many trials are in the competition?

Solution :

Probability of my winning is 3 out of 7 trials $=\frac{3}{7}$

and probability of my friend's winning is $=1-\frac{3}{7}=\frac{4}{7}$

Let there be n numbers of trials.

Therefore, my score after n trials $=\frac{3n}{7}$

and my friend's score after n trials $=\frac{4n}{7}$

Now, the number of trials my friend won more than me is

$\frac{4n}{7}-\frac{3n}{7}=\frac{n}{7}$

We know, that my friend won 5 more trials than me so comparing the actual and given we get,

$\frac{n}{7}=5\\n=7\times5=35$

Therefore, total number of trials is 35.

We can also verify our answer by placing value of n,

After 35 trials I will win $\frac{3\times 35}{7}=15$ trials

and my friend will win $\frac{4\times 35}{7}=20$ trials.

Hence, my friend won 5 more trials than me.

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Answer:

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If we compare the p value and the significance level given $\alpha=0.1$ 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 different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=669$ represent the sample mean

$s=732$ represent the sample standard deviation

$n=1700$ sample size

$\mu_o =68$ represent the value that we want to test

$\alpha=0.$ 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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:

Null hypothesis: $\mu = 634$

Alternative hypothesis: $\mu \neq 634$

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:

$t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971$

P-value

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

$df=n-1=1700-1=1699$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(1699)}>1.971)=0.049$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ 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 different from 634 at 10% of signficance.

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