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yKpoI14uk [10]
3 years ago
14

Suppose you are testing the null hypothesis that a population mean is less than or equal to 46, against the alternative hypothes

is that the population mean is greater than 46. The sample size is 25 and alpha =.05. If the sample mean is 50 and the population standard deviation is 8, the observed z value is:
a) 2.5
b) -2.5
c) 6.25
d) -6.25
e) 12.5
Mathematics
1 answer:
PolarNik [594]3 years ago
6 0

Answer:

Option A) 2.5

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 46

Sample mean, \bar{x} = 50

Sample size, n = 25

Alpha, α = 0.05

Population standard deviation, σ = 8

First, we design the null and the alternate hypothesis

H_{0}: \mu \leq 46\\H_A: \mu > 46

Formula:

z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

Putting all the values, we have

z_{stat} = \displaystyle\frac{50 - 46}{\frac{8}{\sqrt{25}} } =2.5

Option A) 2.5

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\qquad\qquad\huge\underline{{\sf Answer}}♨

As we know ~

Area of the circle is :

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And radius (r) = diameter (d) ÷ 2

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<h3>Problem 1</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 4.4\div 2

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Now find the Area ~

\qquad \sf  \dashrightarrow \: \pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {(2.2)}^{2}

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\qquad \sf  \dashrightarrow \:area  \approx 15.2 \:  \: mm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 2</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 3.7 \div 2

\qquad \sf  \dashrightarrow \:r = 1.85 \:  \: cm

Bow, calculate the Area ~

\qquad \sf  \dashrightarrow \: \pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (1.85) {}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 3.4225 {}^{}

\qquad \sf  \dashrightarrow \:area  \approx 10.75 \:  \: cm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>Problem 3 </h3>

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

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<h3>Problem 4</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

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now, let's calculate area ~

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・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 5</h3>

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