Mathematics, 20.05.2020 12:58 g6birt

Two different samples will be taken from the same population of test scores where the population mean and standard deviation are unknown. The first sample will have 25 data values, and the second sample will have 64 data values. A 95% confidence interval will be constructed for each sample to estimate the population mean. Which number of data values would you expect to give a greater precision (a smaller width) for estimating the population mean? The answer is the sample with data values.

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Answer from: dondre54

The number of data values would you expect to give a greater precision (a smaller width) for estimating the population mean is 64 and this can be determined by using the formula of width.

Given :

The first sample will have 25 data values.The second sample will have 64 data values. 95% confidence interval.

The following steps can be used in order to determine the number of data values would you expect to give a greater precision for estimating the population mean:

Step 1 - The below formula of width is used in order to determine the number of data values.

Step 2 - The formula of width is given below:

$w = 2\times z_{\alpha /2}\times \dfrac{\sigma}{\sqrt{n} }$

where $z_{\alpha /2}$ is the z-score, $\sigma$ is the standard deviation, and 'n' is the sample size.

Step 3 - Now, at n = 25 the formula of 'w' becomes:

$w = 2\times z_{\alpha /2}\times \dfrac{\sigma}{\sqrt{25} }$

$w = \dfrac{1}{5}\left(2\times z_{\alpha /2}\times\sigma }\right)$

Step 4 - Now, at n = 64 the formula of 'w' becomes:

$w = 2\times z_{\alpha /2}\times \dfrac{\sigma}{\sqrt{64} }$

$w = \dfrac{1}{8}\left(2\times z_{\alpha /2}\times\sigma }\right)$

The number of data values would you expect to give a greater precision (a smaller width) for estimating the population mean is 64.

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Answer from: oceaneyez

The sample consisting of 64 data values would give a greater precision.

Step-by-step explanation:

The width of a (1 - α)% confidence interval for population mean μ is:

$\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{n}}$

So, from the formula of the width of the interval it is clear that the width is inversely proportion to the sample size (n).

That is, as the sample size increases the interval width would decrease and as the sample size decreases the interval width would increase.

Here it is provided that two different samples will be taken from the same population of test scores and a 95% confidence interval will be constructed for each sample to estimate the population mean.

The two sample sizes are:

n₁ = 25

n₂ = 64

The 95% confidence interval constructed using the sample of 64 values will have a smaller width than the the one constructed using the sample of 25 values.

Width for n = 25:

$\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{25}}=\frac{1}{5}\ [2\cdot z_{\alpha/2}\cdot \sigma]$

Width for n = 64:

$\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{64}}=\frac{1}{8}\ [2\cdot z_{\alpha/2}\cdot \sigma]$

Thus, the sample consisting of 64 data values would give a greater precision.

Answer from: Quest

the answer is b.

step-by-step explanation:

by the trigonometric ratio tangent, tangent equals opposite divided by adjacent.

therefore, tan(50) equals ac/3

multiply each side by 3 to get: ac=3tan(50)

the answer is b.

Answer from: Quest

step-by-step explanation:

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