Can someone please help me answer this

Legend has it that the great mathematician Carl Friedrich Gauss(1777-1855) at a very young age was told by his teacher to find t

Dima020 [189]

Answer:

6292007867 call

Step-by-step explanation:

4 0

3 years ago

textRequest reports that adults 18–24 years old send and receive 128 texts every day. Suppose we take a sample of 25–34 year old

Butoxors [25]

Testing the hypothesis, we have that:

a)

The null hypothesis is: $H_0: \mu = 128$

The alternative hypothesis is: $H_1: \mu \neq 128$

b) The p-value of the test is of 0.1212.

c) Since the p-value of the test is of 0.1212 > 0.05, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.

d) Since |z| = 1.55 < 1.96, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.

Item a:

At the null hypothesis, we test if the mean is the same, that is, of 128 texts every day, hence:

$H_0: \mu = 128$

At the alternative hypothesis, we test if the mean is different, that is, different of 128 texts every day, hence:

$H_1: \mu \neq 128$

Item b:

We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.

$\mu$ is the value tested at the null hypothesis.

$\sigma$ is the standard deviation of the population.

n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 118.6, \mu = 128, \sigma = 33.17, n = 30$

Hence, the value of the test statistic is:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{118.6 - 128}{\frac{33.17}{\sqrt{30}}}$

$z = -1.55$

Since we have a two-tailed test, as we are testing if the mean is different of a value, the p-value is P(|z| < 1.55), which is 2 multiplied by the p-value of z = -1.55.

Looking at the z-table, z = -1.55 has a p-value of 0.0606

2(0.0606) = 0.1212

The p-value of the test is of 0.1212.

Item c:

Since the p-value of the test is of 0.1212 > 0.05, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.

Item d:

Using a z-distribution calculator, the critical value for a two-tailed test with 95% confidence level is |z| = 1.96.

Since |z| = 1.55 < 1.96, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.

A similar problem is given at brainly.com/question/25369247

4 0

3 years ago

I need the answer to this algebra question. 4 x 38/12 + 2

mylen [45]

Answer: it is 38x/3+2

6 0

2 years ago

Indicar el número entero que corresponde a cada

anyanavicka [17]

Answer:

a)-20

b)+15

c)-3

d)-300

e)-15

f)-5000

g)-400

h)-250

i)+420

3 0

3 years ago

Please man please help me

spayn [35]

Answer:

34.65 dollars

Step-by-step explanation:

multiply the 75 by .45 to get the new price

7 0

2 years ago