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Lubov Fominskaja [6]

3 years ago

12

The null and alternative hypotheses are given. Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed

and the parameter that is being tested. H0: σ = 8.6 H1: σ < 8.6

1 answer:

OlgaM077 [116]3 years ago

3 0

Answer:

Test is Left tailed test

Parameter tested is standard deviation

Step-by-step explanation:

We are given the hypothesis as;

Null hypothesis; H0: σ = 8.6

Alternative hypothesis; H1: σ < 8.6

Where;σ is a constant generally known in statistics as the standard deviation.

Now, it's the alternative hypothesis that will let us know whether this is left tailed, right tailed or two tailed.

Alternative hypothesis says σ < 8.6.

This means that the values of σ that satisfy this hypothesis are less than 8.6 and thus are on the left hand side of 8.6 on a number line. Thus, the shaded region in a normal distribution curve will be on the left.

Thus, it's a left tailed test

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Do parentheses first:

6 +4-2 = 8

Now do brackets:

128/8 = 16

Now multiply:

16 x 8 = 128

The answer is 128

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2 years ago

Suppose that you remain in the store for 15 minutes, and your friend remains in the store for 10 minutes. If each one of you arr

vova2212 [387]

Answer:

P = 0.23375

Step-by-step explanation:

given,

you remain in store for 15 minutes.

you friend remains in store for 10 minutes.

probability that they are store at the same time

from the graph attached below

shaded region shows the required probability,.

probability of the square = 1

probability of the shaded region = 1 - (area of lower triangle - area of upper triangle)

area of lower triangle = \dfrac{1}{2}(0.85)(0.85)

= 0.36125

area of the upper triangle = \dfrac{1}{2}(0.9)(0.9)

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6 0

3 years ago

Suppose a certain study reported that 27.7% of high school students smoke.

Norma-Jean [14]

<u>Given</u>:

Probability of student smoke,

P = 27.7%

= 0.277

Number of students,

n = 632

$q = 1-p$

$=1-0.277$

$=0.723$

(i)

Here,

Number of students (n) = 60

then,

⇒ $n_P=60\times 0.277$

$=16.62$

⇒ $n_q=60\times 0.723$

$=43.38$

We can see that $n_P > 10$ and $n_q>10$ so the normal approximation condition are met.

Now,

$\mu = n_P= 16.62$

$\sigma = \sqrt{n_{Pq}}$

$= \sqrt{60\times 0.277\times 0.723}$

$=3.9664$

Now,

⇒ $P(X$

$=P(Z_{18.5})$

The Z-score is:

= $\frac{18.5-16.62}{3.4664}$

= $0.5423$

hence,

The probability will be:

⇒ $P(Z_{18.5}) = 0.7062$

or,

⇒ $P(Z$

(ii)

Here,

Number of students (n) = 75

$\mu = n_P = 75\times 0.277$

$=20.775$

$\sigma = \sqrt{n_{Pq}}$

$=\sqrt{75\times 0.277\times 0.723}$

$=3.8756$

Now,

⇒ $P(X>17) = P(X> 17.5)$

$=1-P(X \leq 17.5)$

$=1-P(Z_{17.5})$

The Z-score is:

= $\frac{17.5-20.775}{3.8756}$

= $-0.9740$

then, $P(Z_{17.5}) = 0.165$

hence,

The probability will be:

⇒ $P(X>17) = 1-0.165$

$=0.835$

Learn more about probability here:

brainly.com/question/9825651

4 0

3 years ago