Mathematics, 02.09.2020 18:01 Gabby1493

Given the sample mean = 21.15, sample standard deviation = 4.7152, and N = 40 for the low income group, Test the claim that the mean nickel diameter drawn by children in the low income group is greater than 21.21 mm. Test at the 0.1 significance level. Identify the correct alternative hypothesis:p=21.21p=21.21μ>21.21 μ>21.21μ=21.21μ=21.21μ<21.21μ <21.21p<21.21p<21.21p>2 1.21p>21.21

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Answer from: lizettediego10

Null hypothesis:

$\mathtt{H_o : \mu = 21.21}$

Alternative hypothesis

$\mathtt{H_1 : \mu \geq 21.21}$

t = -0.080

Decision Rule: To reject the null hypothesis if t > 1.340 at t

Since t = -0.080, this implies that t < 1.340 that means the t statistics value did not fall into the rejection region. Hence, we fail to reject the null hypothesis at the level of significance 0.10

Conclusion: We conclude that there is insufficient evidence to support the claim that the mean nickel diameter drawn by children in the low-income group is greater than 21.21 mm.

Step-by-step explanation:

Given that:

the sample mean $\overline x$ = 21.15

the standard deviation $\sigma$ = 4.7512

sample size N = 40

The objective is to test the claim that the mean nickel diameter drawn by children in the low-income group is greater than 21.21 mm.

At the level of significance of 0.1

The null hypothesis and the alternative hypothesis for this study can be computed as follows:

Null hypothesis:

$\mathtt{H_o : \mu = 21.21}$

Alternative hypothesis

$\mathtt{H_1 : \mu \geq 21.21}$

This test signifies a one-tailed test since the alternative is greater than or equal to 21.21

The t-test statistics can be computed by using the formula:

$t= \dfrac{\overline x - \mu }{\dfrac{\sigma}{\sqrt{n}}}$

$t = \dfrac{21.15- 21.21 }{\dfrac{4.7152}{\sqrt{40}}}$

$t = \dfrac{-0.06 }{\dfrac{4.7152}{6.3246}}$

t = -0.080

degree of freedom = n - 1

degree of freedom = 40 - 1

degree of freedom = 39

From the t statistical tables,

at the level of significance ∝ = 0.1 and degree of freedom df = 39, the critical value of $\mathtt{{T_{39,0.10} = 1.304}}$

Decision Rule: To reject the null hypothesis if t > 1.340 at t

Since t = -0.080, this implies that t < 1.340 that means the t statistics value did not fall into the rejection region. Hence, we fail to reject the null hypothesis at the level of significance 0.10

Conclusion: We conclude that there is insufficient evidence to support the claim that the mean nickel diameter drawn by children in the low-income group is greater than 21.21 mm.

Answer from: Quest

Cthe residual plot for model 1 has a non random pattern and is a good fit for the data
Ineed an answer ! the table below shows the relationship between the diameter, x, in inches, and th

Ineed an answer ! the table below shows the relationship between the diameter, x, in inches, and th

Answer from: Quest

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step-by-step explanation:

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Given the sample mean = 21.15, sample standard deviation = 4.7152, and N = 40 for the low income gro...

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