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slamgirl [31]
2 years ago
9

Refer to the accompanying data​ table, which shows the amounts of nicotine​ (mg per​ cigarette) in​ king-size cigarettes,​ 100-m

m menthol​ cigarettes, and​ 100-mm nonmenthol cigarettes. The​ king-size cigarettes are​ nonfiltered, while the​ 100-mm menthol cigarettes and the​ 100-mm nonmenthol cigarettes are filtered. Use a 0.05 significance level to test the claim that the three categories of cigarettes yield the same mean amount of nicotine. Given that only the​ king-size cigarettes are not​ filtered, do the filters appear to make a​ difference?

Mathematics
2 answers:
scoundrel [369]2 years ago
7 0

Answer:

So i'm going to assume that the rest of the question is the same as my homework.

It says to determine the null and alternative hypothesis

H0 is mean1=mean2=mean3

H1 is at least one of the three population mean is different from the others

It says find the F test stat

You need some form of a TI-84 plus calculator. From there, you take the data from the King-Size nicotine (mg) in List 1. 100mm-Menthol in List 2. 100-mm nonmenthol in List 3. From there you go to STAT then hit TEST then scroll all the way to the bottom where it says AVOVA. Click enter, from there you see ANOVA(, from there you have to tell it where the lists are. after ANOVA(, put L1, L2,L3 (assuming that's where you put your data into) so when all said and done in your calculator, it should look something like this: ANOVA(L1, L2, L3) then hit enter. The first line where it says F=, that's your test stat.

Find the P value:

You find the P value in the same menu as your test stat. It should be the number right below your test stat. P=

What is the conclusion for this hypothesis ?

Since our P value is less then alpha, we reject the null hypothesis (H0). From there we can conclude that there IS SUFFICIENT evidence to warrant rejection of the claim that the three categories of cigarettes yield the same mean amount of nicotine

Do the filters appear to make a difference?

Given that the king-size cigarettes have the largest mean, it appears that the filters do make a difference

Step-by-step explanation:

liq [111]2 years ago
6 0
This question does not make any sense come back again later
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2 years ago
Read 2 more answers
Domain and range of the function and it’s inverse
sineoko [7]

Answer:

B (second table in list)

Step-by-step explanation:

The function is a quadratic with vertex and y-intercept at (0,-2). The parabola opens up meaning only y - values above -2 are a part of the function. Since range is the set of all y values, then the range is y\geq -2.

The domain is not restricted and it is all real numbers.

The inverse of the function is where x and y are switched. So the range becomes the domain and vice verse.

This means the domain of the inverse is x\geq -2 and the range of the inverse is all real numbers.

8 0
3 years ago
A sample of 11 students using a Ti-89 calculator averaged 31.5 items identified, with a standard deviation of 8.35. A sample of
lina2011 [118]

Answer:

We reject H₀, with CI = 90 % we can conclude that the mean numbers of times identified with Ti-84 does not exceed that of the Ti-89 by more than 1,80

Step-by-step explanation:

Ti-89 Calculator

Sample mean     x = 31,5

Sample standard deviation   s₂  = 8,35

Sample size        n₂ = 11

Ti-84 Calculator

Sample mean     y = 46,2

Sample standard deviation   s₁  = 9,99

Sample size        n₁ = 12

t(s)  = ( y - x - d ) / √s₁²/n₁ + s₂²/n₂

t(s)  = 12,9 / √(99,8/12) + (69,72/11)

t(s)  = 12,9 / √8,32 + 6,34

t(s)  = 12,9 / 3,83

t(s) = 3,37

Test Hypothesis

Null Hypothesis               H₀      y - x > 1,80

Altenative Hypothesis       Hₐ      y - x ≤ 1,80

We have a t(s) = 3,37

We need to compare with t(c)   critical value for

t(c) α; n₁ +n₂-2            df = 12 +11 -2     df  = 21

If we choose  CI = 95 %    then  α = 5 %   α = 0,05

From t-student table

t(c) = 1,72

t(s) = 3,37

t(s) >t(c)

t(s) is in the rejection region therefore we accept Hₐ with CI = 95 %

8 0
2 years ago
The mean number of hours of study time per week for a sample of 562 students is 23. If the margin of error for the population me
Vadim26 [7]

Answer:

The 98% confidence interval for the mean number of hours of study time per week for all students is (20.9, 25.1).

Step-by-step explanation:

Confidence interval:

Sample mean plus/minus the margin of error.

In this question:

Mean of 23.

Margin of error 2.1.

Then

23 - 2.1 = 20.9

23 + 2.1 = 25.1

The 98% confidence interval for the mean number of hours of study time per week for all students is (20.9, 25.1).

3 0
2 years ago
Please help me with these.
frozen [14]

18. The perimeter is simply √3 + √3 + √3 + √3 or 4√3cm, since the perimeter is just all sides added together. You could add the decimal numbers together using a calculator, which I'm not sure if you're supposed to do in your class.

The area is just width times length, so √3 • √3 = 3cm².

19. The perimeter is 2√5 + 2(9 - √5).

This can also be written as 2√5 + 18 - 2√5, which leaves you with a perimeter of 18ft.

The area would be √5 • (9 - √5), which leaves you with (9√5 - 5)ft².

20. The formula for the perimeter (or circumference) of a circle is π times the diameter of the circle. Using the radius of the circle, 1/π, the diameter is 2/π, so

π • 2/π = 2. The circumference of the circle is 2 inches.

The area of the circle is calculated with the equation πr², so

π(1/π)² = π • 1/(π²) = π/(π²) = π. The area is simply π in².

8 0
3 years ago
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