Covert 5 kilograms into grams

Congress regulates corporate fuel economy and sets an annual gas mileage for cars. A company with a large fleet of cars hopes to

attashe74 [19]

Answer:

a) Null hypothesis: $\mu \leq 30.2$

Alternative hypothesis: $\mu > 30.2$

b) $X \sim N(\mu=32.12, \sigma=4.83)$

And the distribution for the random sample is given by:

$\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)$

c) $t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81$

$p_v =P(t_{(49)}>2.81)=0.0035$

d) If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.

e) The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

$\bar X=32.12$ represent the sample mean

$s=4.83$ represent the sample standard deviation

$n=50$ sample size

$\mu_o =30.2$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

a. Define the parameter and state the hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 30.2 mpg, the system of hypothesis would be:

Null hypothesis: $\mu \leq 30.2$

Alternative hypothesis: $\mu > 30.2$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

b. Define the sampling distribution (mean and standard deviation).

Let X the random variable who represent the variable of interest. And we know that the distribution for X is:

$X \sim N(\mu=32.12, \sigma=4.83)$

And the distribution for the random sample is given by:

$\bar X \sim N(\mu=32.12, \frac{\sigma}{\sqrt{n}}=\frac{4.83}{\sqrt{50}}=0.683)$

c. Perform the test and calculate P-value

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{32.12-30.2}{\frac{4.83}{\sqrt{50}}}=2.81$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=50-1=49$

Since is a one right tailed test the p value would be:

$p_v =P(t_{(49)}>2.81)=0.0035$

d. State your conclusion.

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly higher than 30.2.

e. Explain what the p-value means in this context.

The p-value is the probability of obtaining the observed results of a test, assuming that the null hypothesis is correct. And for this case is a value to accept or reject the null hypothesis.

6 0

3 years ago

Anbody know the answer??

LiRa [457]

I'm going to have to go with choice "c". 10x - 1 with the remainder of 5

8 0

3 years ago

Convert Зcis 180° to rectangular form. А. -3 в. -3i с. з D. Зі.

Sunny_sXe [5.5K]

Answer:

-3

Step-by-step explanation:

$3 cis(180^\circ)$ means $3(\cos(180^\circ)+i \sin(180^\circ))$

What is the $x$ -coordinate value that corresponds to $\theta=180^\circ$ . That value is -1.

What is the $y$ -coordinate value that corresponds to $\theta=180^\circ$ . That value is 0.

So this implies $\cos(180^\circ)=-1 \text{ and } \sin(180^\circ)=0$ .

$3 cis(180^\circ)$

$3(\cos(180^\circ)+i \sin(180^\circ))$

$3(-1+i (0))$

$3(-1+0)$

$3(-1)$

$-3$

5 0

3 years ago

20 POINTS

victus00 [196]

Ni is not correct. To solve the equivalent quarterly interest rate, the annual interest rate should be multiplied by the correct ratio. Since the annual interest rate is 4% per year. So in 1 quarter is equal to 0.25 year.

(4% / year) (0.25 year/ 1 quarter) = 1% per quarter

8 0

3 years ago

Which of the following is true?

VARVARA [1.3K]

Answer:

The correct option is c which is if this test was one-tailed instead of two-tailed, you would reject the null.

Step-by-step explanation:

a: This statement cannot be true as the p-value for a 1 tailed test is dependent on the level of significance and other features.

b: This statement cannot be true as there is no valid mathematical correlation between the p-value of the one-tailed test and the current p-value.

c: This statement is true because due to the enhanced level of significance, the null hypothesis will not be rejected.

d: This statement is inverse of statement c which cannot be true.

e: The statement cannot be true as there is no correlation between the current p-value and the p-value of 1 tailed test. The correlation exists between the values of one-tailed and two-tailed p-values.

8 0

3 years ago