Math / Data & Statistics

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Here are summary statistics for the weights of Pepsi in randomly selected cans: $n=36, \bar{x}=0.82409 \mathrm{lb}, s=0.00567 \mathrm{lb}$ . Use a confidence level of $90 \%$ to complete parts (a) through (d) below. a. Identify the critical value $t_{\alpha / 2}$ used for finding the margin of error. $t_{\alpha / 2}=1.69$ (Round to two decimal places as needed.) b. Find the margin of error. $\mathrm{E}=\square \mathrm{lb}$ (Round to five decimal places as needed.) an example Get more help - Clear all Check answer

Studdy Solution

STEP 1

1. The sample size $n = 36$ is large enough to use the $t$ -distribution.
2. The sample mean $\bar{x} = 0.82409$ lb and sample standard deviation $s = 0.00567$ lb are given.
3. The confidence level is $90\%$ , which implies $\alpha = 0.10$ .
4. The critical value $t_{\alpha/2} = 1.69$ is provided.

STEP 2

1. Verify the critical value $t_{\alpha/2}$ .
2. Calculate the margin of error $E$ .

STEP 3

Verify the critical value $t_{\alpha/2}$ for a $90\%$ confidence level with $n-1 = 35$ degrees of freedom. The provided value is $t_{\alpha/2} = 1.69$ .

STEP 4

Calculate the margin of error $E$ using the formula:
$E = t_{\alpha/2} \times \frac{s}{\sqrt{n}}$
Substitute the given values:
$E = 1.69 \times \frac{0.00567}{\sqrt{36}}$

STEP 5

Calculate the standard error:
$\frac{s}{\sqrt{n}} = \frac{0.00567}{6} = 0.000945$

STEP 6

Calculate the margin of error:
$E = 1.69 \times 0.000945 = 0.00159645$
Round to five decimal places:
$E = 0.00160$
The margin of error is:
$\boxed{0.00160} \, \text{lb}$

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Studdy Solution

STEP 1

1. The sample size $n = 36$ is large enough to use the $t$ -distribution.
2. The sample mean $\bar{x} = 0.82409$ lb and sample standard deviation $s = 0.00567$ lb are given.
3. The confidence level is $90\%$ , which implies $\alpha = 0.10$ .
4. The critical value $t_{\alpha/2} = 1.69$ is provided.

STEP 2

1. Verify the critical value $t_{\alpha/2}$ .
2. Calculate the margin of error $E$ .

STEP 3

Verify the critical value $t_{\alpha/2}$ for a $90\%$ confidence level with $n-1 = 35$ degrees of freedom. The provided value is $t_{\alpha/2} = 1.69$ .

STEP 4

Calculate the margin of error $E$ using the formula:
$E = t_{\alpha/2} \times \frac{s}{\sqrt{n}}$
Substitute the given values:
$E = 1.69 \times \frac{0.00567}{\sqrt{36}}$

STEP 5

Calculate the standard error:
$\frac{s}{\sqrt{n}} = \frac{0.00567}{6} = 0.000945$

STEP 6

Calculate the margin of error:
$E = 1.69 \times 0.000945 = 0.00159645$
Round to five decimal places:
$E = 0.00160$
The margin of error is:
$\boxed{0.00160} \, \text{lb}$

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Studdy Solution

STEP 1

STEP 2

1. Verify the critical value tα/2 t_{\alpha/2} tα/2​.2. Calculate the margin of error E E E.

STEP 3

Verify the critical value tα/2 t_{\alpha/2} tα/2​ for a 90% 90\% 90% confidence level with n−1=35 n-1 = 35 n−1=35 degrees of freedom. The provided value is tα/2=1.69 t_{\alpha/2} = 1.69 tα/2​=1.69.

STEP 4

Calculate the margin of error E E E using the formula:E=tα/2×sn E = t_{\alpha/2} \times \frac{s}{\sqrt{n}} E=tα/2​×n​s​Substitute the given values:E=1.69×0.0056736 E = 1.69 \times \frac{0.00567}{\sqrt{36}} E=1.69×36​0.00567​

STEP 5

Calculate the standard error:sn=0.005676=0.000945 \frac{s}{\sqrt{n}} = \frac{0.00567}{6} = 0.000945 n​s​=60.00567​=0.000945

STEP 6

Calculate the margin of error:E=1.69×0.000945=0.00159645 E = 1.69 \times 0.000945 = 0.00159645 E=1.69×0.000945=0.00159645Round to five decimal places:E=0.00160 E = 0.00160 E=0.00160 The margin of error is:0.00160 lb \boxed{0.00160} \, \text{lb} 0.00160​lb

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1. Verify the critical value $t_{\alpha/2}$ .
2. Calculate the margin of error $E$ .

Verify the critical value $t_{\alpha/2}$ for a $90\%$ confidence level with $n-1 = 35$ degrees of freedom. The provided value is $t_{\alpha/2} = 1.69$ .

Calculate the margin of error $E$ using the formula:
$E = t_{\alpha/2} \times \frac{s}{\sqrt{n}}$
Substitute the given values:
$E = 1.69 \times \frac{0.00567}{\sqrt{36}}$

Calculate the standard error:
$\frac{s}{\sqrt{n}} = \frac{0.00567}{6} = 0.000945$

Calculate the margin of error:
$E = 1.69 \times 0.000945 = 0.00159645$
Round to five decimal places:
$E = 0.00160$
The margin of error is:
$\boxed{0.00160} \, \text{lb}$