How do I talk with a live tutoring I took a subscription out

2. Now let's calculate the probability of any two - word phrase in specific order from those nine in the dictionary. We should remember that the probability of occurrence of two or more statistically independent events is equal to the product of their individual probabilities. So,

1/9 * 1/9 = 1/81

3. Using the same Multiplication Rule, we can calculate the probability of a random generation of the phrase "three blind mice", in that specific order. Because there are other phrases that could be generated with those three words, but in different order. The question was specific about the order.

1/9 * 1/9 * 1/9 = 1/729

<u>The probability of randomly generating the phrase "three blind mice" is 1/729 or 0.137%</u>

7 0

3 years ago

Helpppp meee pleaseee

miss Akunina [59]

B= the square root of 65

Formula: a^2= c*2-b^2

a^2= 81-16= 65

Once you find the square root of 65, thats your answer, but here they want you to keep it as 65.

Hope this helps!

5 0

3 years ago

In a recent poll, 778 adults were asked to identify their favorite seat when they fly, and 492 of them chose a window seat. Use

seropon [69]

Answer:

Null hypothesis: $p \leq 0.5$

Alternative hypothesis: $p > 0.5$

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

$p_v =P(Z>7.36)=9.19x10^{-14}$

Since the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of adults prefer window seats when they fly is significantly higher than 0.5 .

Step-by-step explanation:

1) Data given and notation

n=778 represent the random sample taken

X=492 represent the people that chose a window seat.

$\hat p=\frac{492}{778}=0.632$ estimated proportion of people that chose a window seat.

$p_o=0.5$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the majority of adults prefer window seats when they fly:

Null hypothesis: $p \leq 0.5$

Alternative hypothesis: $p > 0.5$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.632 -0.5}{\sqrt{\frac{0.5(1-0.5)}{778}}}=7.36$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(Z>7.36)=9.19x10^{-14}$

5) Conclusion

Since the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of adults prefer window seats when they fly is significantly higher than 0.5 .

6 0

3 years ago

How do I talk with a live tutoring I took a subscription out ​

How do I talk with a live tutoring I took a subscription out