A sample of 10 dips was selected. Assume that the population distribution of dips is normal. The...

Question:

A sample of 10 dips was selected. Assume that the population distribution of dips is normal. The weight of each dip was then recorded. The mean weight was 0.8 oz with a standard deviation of 0.1. The population standard deviation is known to be 0.05 pounds.

1. Identify the following:

(a) {eq}\bar x {/eq}.

(b) {eq}\sigma {/eq}.

(c) n.

2. Which distribution should you use for this problem?

3. Construct a 95% confidence interval for the population mean weight of dips. State the confidence interval and calculate the error bound.

4. Construct a 92% confidence interval for the population mean weight of dips. State the confidence interval and calculate the error bound.

Confidence Interval:

Confidence interval list two values that population parameter is most likely to be contained at given level of confidence. The length of the interval is determined by sample size, sample variability and level of confidence.

Answer and Explanation:

1).

a) {eq}\bar X=0.8 {/eq}


b). {eq}\sigma=0.05 {/eq}


c). {eq}n=10 {/eq}


2).

Standard normal distribution. This is because the it comes from a normal population with known standard deviation.


3).

The upper and lower bounds of confidence interval occurs plus and minus margin of error from the point estimate (sample mean):

{eq}\bar X\pm E {/eq}

Margin of error is calculated by multiplying critical value with standard error of sampling:

{eq}\displaystyle E=Z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt {n}} {/eq}

Find critical value z that correspond to 95% level of confidence:

{eq}\displaystyle \frac{\alpha}{2}=\frac{1-0.95}{2}=0.025\\z_{0.025}=\pm 1.96 {/eq}

Calculate the margin of error:

{eq}\displaystyle E=\pm 1.96\times \frac{0.05}{\sqrt{10}}=\pm 0.31 {/eq}

Construct the interval:

{eq}(0.8\pm 0.31)\\(0.49, 1.11) {/eq}


4).

Repeat the same procedure as in (3). above:

{eq}\displaystyle \frac{\alpha}{2}=\frac{1-0.92}{2}=0.04\\z_{0.04}=\pm 1.75\\\displaystyle \left(0.8\pm 1.75\times \frac{0.05}{\sqrt{10}}\right)\\(0.8\pm 0.028)\\(0.772, 0.828) {/eq}


Learn more about this topic:

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Finding Confidence Intervals with the Normal Distribution

from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 3
14K

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