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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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