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Art [367]
3 years ago
15

Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify t

he null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim. A coin mint has a specification that a particular coin has a mean weight of 2.5 g. A sample of 35 coins was collected. Those coins have a mean weight of 2.49546 g and a standard deviation of 0.01839 g. Use a 0.05 significance level to test the claim that this sample is from a population with a mean weight equal to 2.5 g. Do the coins appear to conform to the specifications of the coin​ mint?
Mathematics
1 answer:
lorasvet [3.4K]3 years ago
5 0

Answer:

At a significance level of 0.05, there is not enough evidence to support the claim that the population mean is signficantly different from 2.5 g.

We can not conclude that the sample is drawn from a population with mean different from 2.5 g. This does not confirm that the sample is drawn from a population with mean 2.5 g (we can not confirm the null hypothesis, even if it is failed to be rejected).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean is signficantly different from 2.5 g.

Then, the null and alternative hypothesis are:

H_0: \mu=2.5\\\\H_a:\mu\neq 2.5

The significance level is 0.05.

The sample has a size n=35.

The sample mean is M=2.49546.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.01839.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.01839}{\sqrt{35}}=0.0031

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.49546-2.5}{0.0031}=\dfrac{0}{0.0031}=-1.4605

The degrees of freedom for this sample size are:

df=n-1=35-1=34

This test is a two-tailed test, with 34 degrees of freedom and t=-1.4605, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t

As the P-value (0.1533) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the population mean is signficantly different from 2.5 g.

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1. A company has a cash portfolio measured in millions. The drift is 0.1 per month, variance is 0.16per month. The initial cash
Romashka-Z-Leto [24]

Answer:

Step-by-step explanation:

From the information given:

The probability distribution at the end of 6 months is determined as follows:

After 6 months;

Mean of probability distribution =  value of Initial cash  + \alphaT  

=2.0 +(0.1 × 6)

=2.6

After 6 months;

The probability distribution's standard deviation is estimated by using the following formula:

Standard deviation:

=  b\sqrt{T}

= 0.4 \times \sqrt{6}

= 0.9798

Hence, after 6 months;

The company's cash position is supposed to be allocated monthly, with the following expenses.

Mean                            2.6

Standard deviation     0.9798

Variance                      0.96

After 12 months, the probability distribution is as follows:

Mean = value of Initial cash  + \alphaT

= 2.0 +(0.1 × 12)

= 3.2

The standard deviation is:

The standard deviation of probability distribution = b \sqrt{T}

= 0.4 \times \sqrt{12}

= 1.3856

Hence, after 6 months;

The company's cash position is supposed to be allocated monthly, with the following expenses.

Mean                            3.2

Sandard deviation      1.3856

Variance                      1.92

b)  

in 6-month distribution, the probability of the negative value of the cash position is as follows.

Now, for us to find the negative cash distribution;

We need to estimate the z -scores value.

The z-score inform us greatly on the concept of how far a particular data point is from the mean.

For a normal distribution;

z = \dfrac{x-\mu}{\sigma}

Here;

the value of x = zero as a result that if it exceeds zero. the cash position will be negative.

∴

z = \dfrac{x-\mu}{\sigma}

z = \dfrac{0 - 2.6}{0.9798}

z = -2.6536

Using the standard distribution tables, it is now possible to calculate that the likelihood N(-2.65) equals 0.004 or 0.4 percent.

As a result, there's a 0.4 percent chance of getting a negative cash balance after six months.

For 12 months distribution:

The Probability of negative cash position is calculated as follows:

z = \dfrac{x-\mu}{\sigma} \\ \\ z = \dfrac{0-3.2}{1.3856} \\ \\ z = -2.3094

Using the standard distribution tables,

N(-2.31) equals 0.0104 or 1.04 percent.

As a result, there's a 1.04 percent chance of getting a negative cash balance after 1 year  

c) To determine the time period over which the likelihood of achieving a negative cash condition is highest, it's necessary to examine the z-score more closely. Essentially, the z-score measures the difference between a given value(x) and the mean of all potential values (\mu), expressed in terms of the total set's standard deviation (\sigma)

This suggests that the higher the z-score, the greater the difference occurring between x and \mu, and thus the likelihood of receiving x is minimal. As a result, the best chance of finding a certain value is when the z-score is the lowest.

To do so, calculate the derivative of the z-score in relation to the time interval. The point where the derivative is equivalent to zero is where the z-scores are at their lowest.

The first step is to go over the z-score formula in more detail, as seen below.;

z = \dfrac{x-\mu}{\sigma} \\ \\ z = \dfrac{0-(initial \ value + \alpha T)}{b \sqrt{T}} \\ \\ z = \dfrac{-initial \ value }{b\sqrt{T}}-\dfrac{a \sqrt{T}}{b} \\ \\

Now, compute the derivative of this equation with respect to T as follows:

\dfrac{dz}{dT}= \dfrac{initial \ value \times T^{-\dfrac{3}{2}}}{2b} - \dfrac{aT^{-\dfrac{1}{2}}}{2b}

Now, figure out the value of T at which this derivative is equal to zero by substituting all values as follows:

0 = \dfrac{2.0 \times T^{-\dfrac{3}{2}}}{2\times 0.4}- \dfrac{0.1 \times T^{-\dfrac{1}{2}}}{2 \times 0.4} \\ \\  \\ 0.1 \times T^{-\dfrac{1}{2}}= 2.0 \times T^{-\dfrac{3}{2}} \\ \\  \\T = \dfrac{2}{0.1} \\ \\ \\ T = 20

As a result, the time period in which achieving a negative cash condition is = 20 months.

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Step-by-step explanation:

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