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frez [133]
3 years ago
13

For an analysis of variance comparing three treatment means, H0 states that all three population means are the same and H1 state

s that all three population means are different.
A. True
B. False
Mathematics
1 answer:
antoniya [11.8K]3 years ago
7 0

Answer:

False

Step-by-step explanation:

The analysis of variance may be described as an hypothesis test which is used to make comparison between variables of two or more independent groups. The null hypothesis is always of the notion that there is no difference in the means. While the alternative hypothesis is the opposite, for two independent groups, the alternative hypothesis is that both means are different, or not equal or not the same. However. When we have more than 2 independent groups, then the alternative hypothesis is stated as : 'the means are not all equal'. This means that the means of each group does not all have to be different, but the mean of one group may be different from that of the other groups or the mean of two groups are different from the other groups and so on.

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Simplify (1-√3)(⅓+√3) leaving your answer in the form p+q√3​
Mars2501 [29]

Answer:

\frac{-8 +2\sqrt{3} }{3}

Step-by-step explanation:

When working with surds we need to take note of the roots present there.

To expand this equation we can do it the following way noting that √3 X √3 = 3

<em></em>

<em>Expanding (1-√3)(⅓+√3)</em>

1 X 1/3 = 1/3

1 X √3 = √3

-√3 X 1/3 =-√3/3

√3 X √3 = 3

hence, expanding the equation, we have

1/3 + √3 -√3/3 + 3

We can simply group the like terms and add them up.

[1/3 +3] +[√3-√3/3]

10/3 + \frac{2\sqrt{3} }{3}

= \frac{-8 +2\sqrt{3} }{3}

8 0
3 years ago
The numbers 81, 45,27 are multiplies of
malfutka [58]
They are multiplis of 9
4 0
3 years ago
Read 2 more answers
A box contains 24 transistors,4 of which are defective. If 4 are sold at random,find the following probabilities. i. Exactly 2 a
zavuch27 [327]

SOLUTION

This is a binomial probability. For i, we will apply the Binomial probability formula

i. Exactly 2 are defective

Using the formula, we have

\begin{gathered} P_x=^nC_x\left(p^x\right?\left(q^{n-x}\right) \\ Where\text{ } \\ P_x=binomial\text{ probability} \\ x=number\text{ of times for a specific outcome with n trials =2} \\ p=\text{ probability of success = }\frac{4}{24}=\frac{1}{6} \\ q=probability\text{ of failure =1-}\frac{1}{6}=\frac{5}{6} \\ ^nC_x=\text{ number of combinations = }^4C_2 \\ n=\text{ number of trials = 4} \end{gathered}

Note that I made the probability of being defective as the probability of success = p

and probability of none defective as probability of failure = q

Exactly 2 are defective becomes the binomial probability

\begin{gathered} P_x=^4C_2\times\lparen\frac{1}{6})^2\times\lparen\frac{5}{6})^{4-2} \\ P_x=6\times\frac{1}{36}\times\frac{25}{36} \\ P_x=\frac{25}{216} \\ =0.1157 \end{gathered}

Hence the answer is 0.1157

(ii) None is defective becomes

\begin{gathered} \lparen\frac{5}{6})^4=\frac{625}{1296} \\ =0.4823 \end{gathered}

hence the answer is 0.4823

(iii) All are defective

\begin{gathered} \lparen\frac{1}{6})^4=\frac{1}{1296} \\ =0.00077 \end{gathered}

(iv) At least one is defective

This is 1 - probability that none is defective

\begin{gathered} 1-\lparen\frac{5}{6})^4 \\ =1-\frac{625}{1296} \\ =\frac{671}{1296} \\ =0.5177 \end{gathered}

Hence the answer is 0.5177

3 0
1 year ago
The meat department of a local supermarket packages ground beef using meat trays of two sizes: 1 designed to hold 1 lb of meet a
r-ruslan [8.4K]

Answer:

c) 0.932

99% confidence interval for average weights of all packages sold in small meat trays.

(0.932 ,1.071)

Step-by-step explanation:

Explanation:-

Given random sample of 35 packages in small meat trays produced weight with an average of 1.01 lbs. and standard deviation  of 0.18 lbs.

size of the sample 'n' = 35

mean of the sample x⁻= 1.01lbs

standard deviation of the sample 'S' = 0.18lbs

<u>The 99% confidence intervals are given by</u>

(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-} +t_{\alpha } \frac{S}{\sqrt{n} } )

The degrees of freedom γ=n-1 =35-1=34

tₐ =  2.0322

99% confidence interval for average weights of all packages sold in small meat trays

(1.01 - 2.0322 \frac{0.18}{\sqrt{35} } , 1.01+2.0322 \frac{0.18}{\sqrt{35} } )

( 1.01 - 0.06183 , 1.01+0.06183)

(0.932 ,1.071)

<u>Final answer</u>:-

<u>99% confidence interval for average weights of all packages sold in small meat trays.</u>

<u>(0.932 ,1.071)</u>

5 0
3 years ago
Andrei's grandfather offered to give him a gift at 50% of the amount of money he saved in one year. Andrei saved $120 dollars. H
Dafna1 [17]

Answer:

$60

Step-by-step explanation:

$120/2(half) = $60

3 0
3 years ago
Read 2 more answers
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