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Tasya [4]
4 years ago
12

A magazine article states that the mean weight of one-year-old boys is the same as that of one-year-old girls. Does the confiden

ce interval contradict this statement? The confidence interval this statement

Mathematics
1 answer:
kirill115 [55]4 years ago
4 0

Answer:

Yes, the confidence interval contradict this statement.

Step-by-step explanation:

The complete question is attached below.

The data provided is:

n_{1}=318\\n_{2}=297\\\bar x_{1}=25\\\bar x_{2}=24.1\\s_{1}=3.6\\s_{2}=3.8

Since the population standard deviations are not provided, we will use the <em>t</em>-confidence interval,

CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\cdot s_{p}\cdot\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}

Compute the pooled standard deviation as follows:

s_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}=\sqrt{\frac{(318-1)(3.6)^{2}+(297-1)(3.8)^{2}}{318+297-2}}=2.9723

The critical value is:

t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (318+297-2)}=t_{0.025, 613}=1.962

*Use a <em>t</em>-table.

The 95% confidence interval is:

CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\cdot s_{p}\cdot\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}

     =(25-24.1)\pm 1.962\times 2.9723\times \sqrt{\frac{1}{318}+\frac{1}{297}}\\\\=0.90\pm 0.471\\\\=(0.429, 1.371)\\\\\approx (0.43, 1.37)

The 95% confidence interval for the difference between the mean weights is (0.43, 1.37).

To test the magazine's claim the hypothesis can be defined as follows:

<em>H</em>₀: There is no difference between the mean weight of 1-year old boys and girls, i.e. \mu_{1}-\mu_{2}=0.

<em>H</em>ₐ: There is a significant difference between the mean weight of 1-year old boys and girls, i.e. \mu_{1}-\mu_{2}\neq 0.

Decision rule:

If the confidence interval does not consists of the null value, i.e. 0, the null hypothesis will be rejected.

The 95% confidence interval for the difference between the mean weights does not consists the value 0.

Thus, the null hypothesis will be rejected.

Conclusion:

There is a significant difference between the mean weight of 1-year old boys and 1-year old girls.

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The unit of time that will make the statement true is "day"

<h3>What unit of time will make the statement true?</h3>

The conversion equation is given as:

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a) The minimum head breadth that will fit the clientele = 4.105 inches to 3d.p = 4.1 inches to 1 d.p

b) The maximum head breadth that will fit the clientele = 8.905 inches to 3 d.p = 8.9 inches to 1 d.p

Step-by-step explanation:

This is normal distribution problem.

A normal distribution has all the data points symmetrically distributed around the mean in a bell shape.

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Using the table to obtain the value of z''

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P(z ≥ 1.995) = 0.023

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z'' = (x - xbar)/σ

1.995 = (x - 6.1)/1

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