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uranmaximum [27]

3 years ago

7

A sample of size 168, taken from a normally distributed population whose standard deviation is known to be 8.60, has a sample me

an of 80.60. Suppose that we have adopted the null hypothesis that the actual population mean is equal to 80, that is, H0 is that μ = 80 and we want to test the alternative hypothesis, H1, that μ ≠ 80, with level of significance α = 0.1. The upper limit of a 95% confidence interval for the population mean would equal: _______

1 answer:

denis23 [38]3 years ago

7 0

Answer:

The 95% confidence interval for the population mean is

(79.2996, 81.9004)

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given that the sample size 'n' = 168</em>

<em>Let 'X' be a Random variable in a normal distribution </em>

<em>Given that mean of the sample x⁻ = 80.60</em>

Given that the standard deviation of the Population = 8.60

<u><em>Step(ii):-</em></u>

<em>The 95% confidence interval for the population mean is determined by</em>

<em> </em> $(x^{-} - Z_{0.05} \frac{S.D}{\sqrt{n} } ,x^{-} + Z_{0.05} \frac{S.D}{\sqrt{n} } )$ <em></em>

<em></em> $(80.60 -1.96 \frac{8.60}{\sqrt{168} } ,80.60 + 1.96 \frac{8.60}{\sqrt{168} } )$ <em></em>

(80.60 -1.3004 , 80.60+1.3004)

(79.2996 , 81.9004)

<u><em>Final answer:-</em></u>

<em>The 95% confidence interval for the population mean is</em>

<em>(79.2996, 81.9004)</em>

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What is the least common multiple of 3 and 9

Tju [1.3M]

LCM(3, 9) = 9

because 9 = 3 × 3

5 0

3 years ago

Read 2 more answers

Please helpppppp mee thanks everyone

netineya [11]

Answer:

$\frac{32}{35}$

Step-by-step explanation:

To answer this question you need to get the same denominator (bottom number of a fraction ) on both values

Because 7 is a factor of 35, you can multiply the whole fraction by 5, (so that the first fraction has a denominator of 35 which is the same as the other value)

$\frac{4}{7}$ ×5 $= \frac{20}{35}$

then you add them up but remember if the denominator is the same you do <u>not </u> add them.

$\frac{20}{35} +\frac{12}{35} = \frac{32}{35}$

Hope the answer is right, Ask if unsure :)

7 0

2 years ago

Damien has 30 coins, all nickels and dimes. The total value of all these coins is $2.05. How many nickels does he have?

Damm [24]

Hi there!

nickels:19

We know that there is only nickels and dimes in his coin collection..

<u><em>1.Multiply two terms:</em></u>

<u><em /></u> $10$ <u><em>·</em></u> $10=1.00$ <u><em /></u>

<u><em>Then there would be 20 nickels but 1+1=2 not 2.05.</em></u>

<u><em>2.Multiply the terms:</em></u>

<u><em /></u> $11$ <u><em>·</em></u> $10=1.10$ <u><em /></u>

<u><em /></u> $19$ <u><em>·</em></u> $5=0.95$ <u><em /></u>

<u><em>3.Combine the two terms:</em></u>

<u><em /></u> $1.10+0.95=2.05$ <u><em /></u>

Therefore, there would be 19 nickels...

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4 0

3 years ago

If A^2=A, which matrix is matrix A

Ronch [10]

Consider all options:

A.

$\left[\begin{array}{cc}5&5\\-4&-4\end{array}\right] \cdot \left[\begin{array}{cc}5&5\\-4&-4\end{array}\right] =\left[\begin{array}{cc}5\cdot 5+5\cdot (-4)&5\cdot 5+5\cdot (-4)\\-4\cdot 5+(-4)\cdot (-4)&-4\cdot 5+(-4)\cdot (-4)\end{array}\right]=$

$=\left[\begin{array}{cc}5&5\\-4&-4\end{array}\right].$

This option is true.

B.

$\left[\begin{array}{cc}6&5\\5&6\end{array}\right] \cdot \left[\begin{array}{cc}6&5\\5&6\end{array}\right] =\left[\begin{array}{cc}6\cdot 6+5\cdot 5&6\cdot 5+5\cdot 6\\5\cdot 6+6\cdot 5&5\cdot 5+6\cdot 6\end{array}\right]=$

$=\left[\begin{array}{cc}61&60\\60&61\end{array}\right].$

This option is false.

C.

$\left[\begin{array}{cc}0.5&-0.5\\-0.5&0.5\end{array}\right] \cdot \left[\begin{array}{cc}0.5&-0.5\\-0.5&0.5\end{array}\right] =\left[\begin{array}{cc}0.5\cdot 0.5+(-0.5)\cdot (-0.5)&0.5\cdot (-0.5)+(-0.5)\cdot 0.5\\-0.5\cdot 0.5+0.5\cdot (-0.5)&-0.5\cdot (-0.5)+0.5\cdot 0.5\end{array}\right]=$

$=\left[\begin{array}{cc}0.5&-0.5\\-0.5&0.5\end{array}\right].$

This option is true.

D.

$\left[\begin{array}{cc}0.5&0.5\\-0.5&0.5\end{array}\right] \cdot \left[\begin{array}{cc}0.5&0.5\\-0.5&0.5\end{array}\right] =\left[\begin{array}{cc}0.5\cdot 0.5+0.5\cdot (-0.5)&0.5\cdot 0.5+0.5\cdot 0.5\\-0.5\cdot 0.5+0.5\cdot (-0.5)&-0.5\cdot 0.5+0.5\cdot 0.5\end{array}\right]=$

$=\left[\begin{array}{cc}0&0.5\\-0.5&0\end{array}\right].$

This option is false.

E.

$\left[\begin{array}{cc}-6&-6\\5&5\end{array}\right] \cdot \left[\begin{array}{cc}-6&-6\\5&5\end{array}\right] =\left[\begin{array}{cc}-6\cdot (-6)+(-6)\cdot 5&-6\cdot (-6)+(-6)\cdot 5\\5\cdot (-6)+5\cdot 5&5\cdot (-6)+5\cdot 5\end{array}\right]=$

$=\left[\begin{array}{cc}6&6\\-5&-5\end{array}\right].$

This option is false.

Answer: correct options are A and C.

8 0

3 years ago

Read 2 more answers

Let A = (1, 0, 3) and B = (−3, 2, 1).

Romashka-Z-Leto [24]

Answer:

a) $(-\frac{\sqrt{24}}{6} ,\frac{\sqrt{24}}{12} ,-\frac{\sqrt{24}}{12})$

b) $(-1,1,2)$

c) $-2x+y-z=1$

Step-by-step explanation:

a)

AB, OB, OA are vectors.

$AB = OB - OA = (-3, 2, 1) - (1, 0, 3) = (-4, 2, -2)\\||AB|| = \sqrt{(-4)^2+2^2+(-2)^2}=\sqrt{24}\\$

Normalizing AB: $\frac{AB}{||AB||} = (\frac{-4}{\sqrt{24}},\frac{2}{\sqrt{24}}, \frac{-2}{\sqrt{24}})=(-\frac{\sqrt{24}}{6} ,\frac{\sqrt{24}}{12} ,-\frac{\sqrt{24}}{12})$

b)

OM, OA, OB are vectors.

$OM = \frac{1}{2}OA+\frac{1}{2}OB=(\frac{-3+1}{2} ,\frac{2+0}{2}, \frac{1+3}{2} )=(-1,1,2)$

So, $M = (-1,1,2)$

c)

$-4x+2y-2z=(-4)(-1)+(2)(1)+(-2)(2)=4+2-4=2$

Hence,

$-2x+y-z=1$

5 0

3 years ago