In a random sample of 11 residents of the state of Washington, the mean waste recycled per person...

Question:

In a random sample of 11 residents of the state of Washington, the mean waste recycled per person per day was 1.4 pounds with a standard deviation of 0.69 pounds.

a. Determine the 80% confidence interval for the mean waste recycled per person per day for the population of Washington. Assume the population is normally distributed.

b. Find the critical value that should be used in constructing the confidence interval. (Round answer to 3 decimal places.)

Confidence Interval:

In this question, we will use the t distribution to calculate and construct the 80% confidence interval for the mean waste recycled per person per day for the population of Washington. The t distribution is a sampling distribution with (n-1) degree of freedom. An excel function TINV() is used to get the critical value at 20% level of significance.

Answer and Explanation:

Given that,

  • Sample size, {eq}n = 11 {/eq}
  • Sample mean, {eq}\bar{x} = 1.4 {/eq}
  • Sample standard deviation, {eq}s = 0.69 {/eq}


Degree of freedom, {eq}n-1 = 11 - 1 = 10 {/eq}

a)

The 80% confidence interval for the population mean is defined as:

{eq}\bar{x} \pm t_{0.20/2}\times \frac{s}{\sqrt{n}} {/eq}


Excel function for the confidence coefficient at 20%:

=TINV(0.2,10)


Now,

{eq}1.4 \pm 1.3722\times \frac{0.69}{\sqrt{11}}\\ 1.1145 < \mu < 1.6855 {/eq}


The 80% confidence interval for the mean waste recycled per person per day for the population of Washington= (1.1145, 1.6855).

b)

The critical value, {eq}t_{0.20/2,10} = 1.372 {/eq}


Learn more about this topic:

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Using the t Distribution to Find Confidence Intervals

from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 6
6.2K

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