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puteri [66]
4 years ago
7

a random sample of 10 subjects have weight with a standard deviation of 13.1392 kg.what is the variance of their weight ​

Mathematics
1 answer:
Irina18 [472]4 years ago
6 0

Answer:

The variance of their weight σ² = 172.63kg

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given sample size 'n' = 10

Given weight of the standard deviation (σ)  =13.1392

we know that standard deviation (σ)=√ variance

The variance of their weight σ²  =(13.1392)²

                                                σ² = 172.63

The variance of their weight σ² = 172.63kg

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Answer:

Step-by-step explanation:

From the given information, we can compute the table showing the summarized statistics of the two alloys A & B:

                                            Alloy A        Alloy B

Sample mean                    \bar {x} _A = 49.5      \bar {x} _B = 45.5

Equal standard deviation    \sigma_A = 5         \sigma_B= 5

Sample size                          n_A = 30        n_{B}= 30

Mean of the sampling distribution is :

\mu_{\bar{X_1}-\bar{X_2}}= 49.5-45.5  \\ \\ = 4.0

Standard deviation of sampling distribution:

\sigma_{\bar{x_1}-\bar{x_2} }= \sqrt{\sigma^2_{\bar{x_1}-\bar{x_2} }} \\ \\ = \sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma^2_2}{n_2} } \\ \\ =\sqrt{\dfrac{25}{30}+\dfrac{25}{30}} \\\\=\sqrt{1.667}  \\ \\ =1.2909

Hypothesis testing.

Null hypothesis:  H_o: \mu_A -\mu_B = 0

Alternative hypothesis:  H_A:\mu_A -\mu_B > 4

The required probability is:

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This implies that a minimal chance of probability shows that the difference of 4 is not likely, provided that the two population means are the same.

b)

Since the P-value is very small which is lower than any level of significance.

Then, we reject H_o and conclude that there is enough evidence to fully support alloy A.

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