Consider the probability that fewer than 11 out of 149 houses will lose power once a year. Assume the probability that a given h
1 answer:
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Answer:
Step-by-step explanation:
To solve these equations involving variables and exponents we need to follow these steps.
1) We need to find out the factor that is common in the equation.
2) After taking common, solve the equation. We can add or subtract only those values that have same bases.
1)
here we can see, both numbers are divisible by 2, so taking 2 common
It cannot be further simplified because both number donot have same bases.
3.
We can take 4n common
5. -12a -3
Here -3 cam be taken common
= -3(-12a/-3 -3/-3)
= -3(4a +1)
7.
here the smallest power of n is n^3 so, we can take n^3 common and both coefficients are divisible by 4 so taking 4n^3 common
9.
Here we cannot take k common, as k is not a multiple of 10. For taking common it should be divisible by each value in the equation. But each value s divisible by 5 so, taking 5 common
11.
Here we cannot take n common, as n is not a multiple of -60. For taking common it should be divisible by each value in the equation. But each value s divisible by 10 so, taking 10 common
13.
Here we cannot take n common, as n is not a multiple of 28. For taking common it should be divisible by each value in the equation. But each value s divisible by -4 so, taking -4 common
15.
Here we cannot take n common, as n is not a multiple of 18. For taking common it should be divisible by each value in the equation. But each value s divisible by 9 so, taking 9 common
17.
Answer:
We conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.
Step-by-step explanation:
We are given that a study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois.
Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids.
Let = <u><em>proportion of Illinois high school freshmen who have used anabolic steroids.</em></u>
= <u><em>proportion of Illinois high school seniors who have used anabolic steroids.</em></u>
SO, Null Hypothesis, :
{means that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}
Alternate Hypothesis, :
{means that there is a significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}
The test statistics that would be used here <u>Two-sample z test for</u> <u>proportions</u>;
T.S. = ~ N(0,1)
where, = sample proportion of high school freshmen who have used anabolic steroids =
= 0.0203
= sample proportion of high school seniors who have used anabolic steroids =
= 0.0176
= sample of high school freshmen = 1679
= sample of high school seniors = 1366
So, <u><em>the test statistics</em></u> =
= 0.545
The value of z test statistics is 0.545.
Since, in the question we are not given with the level of significance so we assume it to be 5%. <u>Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</u>
Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.
Therefore, we conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.