After unloading 96 pounds of mulch from the truck, there were still 24 pounds

A Local group claims that the police issue more than 60 speeding tickets a day in thire area. To prove their point, they randoml

MissTica

Answer:

The group's claim is not supported at 1% significance level.

Step-by-step explanation:

given that a local group claims that the police issue more than 60 speeding tickets a day in thire area

The data they collected for 2 weeks is given below:

70 48 41 68 69 55 70

57 60 83 32 60 72 58

$\bar x = 60.2\\s = 12.9\\n=14$

Std error of mean = $\frac{12.9}{\sqrt{14} } \\=3.4477$

$H_0: \bar x=60\\H_a: \bar x >60$

(Right tailed test at 1% significance level)

Mean difference = 0.10

Test statistic t with df 13 = mean difference/std error

= 0.029

p value = 0.4886

Since p >0.01, we accept null hypothesis

The group's claim is not supported at 1% significance level.

7 0

2 years ago

The aspect of consistency means that ____. a. the association should be compatible with existing theory and knowledge b. the ass

Mama L [17]

Answer:

b. the association is consistent when results are repeated in studies in different settings using different methods

Step-by-step explanation:

-In statistics and research, internal consistency is typically a measure based on the correlations between different items on the same test (or the same subscale on a larger test).

-It measures whether several items that propose to measure the same general construct produce similar scores.

8 0

2 years ago

IM GIVING BRAINLIEST!

Mariana [72]

Sorry i don’t know the answer but i’m new to this app, and was wondering what brainliest is?:)

6 0

2 years ago

A ball is thrown in the air from a height of 3 m with an initial velocity of 12 m/s. To the nearest tenth of a second, how long

stepladder [879]

Answer: 1.4 seconds

<u>Step-by-step explanation:</u>

The equation is: h(t) = at² + v₀t + h₀ where

a is the acceleration (in this case it is gravity)
v₀ is the initial velocity
h₀ is the initial height

Given:

a = -9.81 (if it wasn't given in your textbook, you can look it up)
v₀ = 12
h₀ = 3

Since we are trying to find out when it lands on the ground, h(t) = 0

EQUATION: 0 = 9.81t² + 12t + 3

Use the quadratic equation to find the x-intercepts

a=-9.81, b=12, c=3

$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(12)\pm \sqrt{(12)^2-4(-9.81)(3)}}{2(-9.81)}\\\\\\x=\dfrac{-12\pm 16.2}{-19.62}\\\\\\x=\dfrac{-12+ 16.2}{-19.62}=-0.2\qquad x=\dfrac{-12- 16.2}{-19.62}=\large\boxed{1.4}\\$

Note: Negative time (-0.2) is not valid

4 0

3 years ago

Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati

Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c. D. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X $\sim$ N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

$P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})$

$P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})$

$P(X < 76) = P(Z< \dfrac{3}{12.5})$

$P(X < 76) = P(Z< 0.24)$

From the standard normal distribution tables,

$P(X < 76) = 0.5948$

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

$P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})$

$P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})$

$P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})$

$P( \overline X < 76) = P(Z< 1.2)$

From the standard normal distribution tables,

$P(\overline X < 76) = 0.8849$

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

In order to determine the probability in part (b); the normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

5 0

2 years ago