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A simple random sample of items resulted in a sample mean of . The population standard deviation is . a. Compute the confidence interval for the population mean. Round your answers to one decimal place. ( , ) b. Assume that the same sample mean was obtained from a sample of items. Provide a confidence interval for the population mean. Round your answers to two decimal places. ( , ) c. What is the effect of a larger sample size on the interval estimate? Larger sample provides a - Select your answer - margin of error.

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MrRoyal

Answer:

(a): The 95% confidence interval is (46.4, 53.6)

(b): The 95% confidence interval is (47.9, 52.1)

(c): Larger sample gives a smaller margin of error.

Step-by-step explanation:

Given

[tex]n = 30[/tex] -- sample size

[tex]\bar x = 50[/tex] -- sample mean

[tex]\sigma = 10[/tex] --- sample standard deviation

Solving (a): The confidence interval of the population mean

Calculate the standard error

[tex]\sigma_x = \frac{\sigma}{\sqrt n}[/tex]

[tex]\sigma_x = \frac{10}{\sqrt {30}}[/tex]

[tex]\sigma_x = \frac{10}{5.478}[/tex]

[tex]\sigma_x = 1.825[/tex]

The 95% confidence interval for the z value is:

[tex]z = 1.960[/tex]

Calculate margin of error (E)

[tex]E = z * \sigma_x[/tex]

[tex]E = 1.960 * 1.825[/tex]

[tex]E = 3.577[/tex]

The confidence bound is:

[tex]Lower = \bar x - E[/tex]

[tex]Lower = 50 - 3.577[/tex]

[tex]Lower = 46.423[/tex]

[tex]Lower = 46.4[/tex] --- approximated

[tex]Upper = \bar x + E[/tex]

[tex]Upper = 50 + 3.577[/tex]

[tex]Upper = 53.577[/tex]

[tex]Upper = 53.6[/tex] --- approximated

So, the 95% confidence interval is (46.4, 53.6)

Solving (b): The confidence interval of the population mean if mean = 90

First, calculate the standard error of the mean

[tex]\sigma_x = \frac{\sigma}{\sqrt n}[/tex]

[tex]\sigma_x = \frac{10}{\sqrt {90}}[/tex]

[tex]\sigma_x = \frac{10}{9.49}[/tex]

[tex]\sigma_x = 1.054[/tex]

The 95% confidence interval for the z value is:

[tex]z = 1.960[/tex]

Calculate margin of error (E)

[tex]E = z * \sigma_x[/tex]

[tex]E = 1.960 * 1.054[/tex]

[tex]E = 2.06584[/tex]

The confidence bound is:

[tex]Lower = \bar x - E[/tex]

[tex]Lower = 50 - 2.06584[/tex]

[tex]Lower = 47.93416[/tex]

[tex]Lower = 47.9[/tex] --- approximated

[tex]Upper = \bar x + E[/tex]

[tex]Upper = 50 + 2.06584[/tex]

[tex]Upper = 52.06584[/tex]

[tex]Upper = 52.1[/tex] --- approximated

So, the 95% confidence interval is (47.9, 52.1)

Solving (c): Effect of larger sample size on margin of error

In (a), we have:

[tex]n = 30[/tex]     [tex]E = 3.577[/tex]

In (b), we have:

[tex]n = 90[/tex]    [tex]E = 2.06584[/tex]

Notice that the margin of error decreases when the sample size increases.

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