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Each child in a sample of 65 low-income children was administered a language and communication...

Question:

Each child in a sample of 65 low-income children was administered a language and communication exam. The sentence complexity scores had a mean of 7.62 and a standard deviation of 8.91.

a. From the sample, estimate the true mean sentence complexity score of all low-income children.

b. Form a 90% confidence interval for the estimate.

Constructing Confidence Interval for the Mean:

Confidence interval is a type of interval estimation that give range within which true population mean is most likely to fall at given level of confidence. The width of the confidence occurs plus and minus margin of error from the point estimate.

Answer and Explanation:

a).

According to sampling distribution of the mean, the sample mean is equal to population mean:

{eq}\mu_{\bar x}=\mu=7.62 {/eq}


b).

We'll construct the confidence interval using student t distribution because population standard deviation is unknown:

{eq}\displaystyle (\bar X\pm t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}) {/eq}

Find the t critical:

  • Calculate the level of significance that correspond to 90% level of confidence:

{eq}\begin{align*} \alpha&=1-CI\\&=1-0.90\\&=0.1\\\displaystyle \frac{\alpha}{2}&=0.05 \end{align*} {/eq}

  • Calculate the degrees of freedom:

{eq}\begin{align*} Df&=n-1\\&=65-1\\&=64 \end{align*} {/eq}

  • Use the software or the t table to find the t critical:

{eq}t_{0.05,df=64}=1.669 {/eq}

Now construct the confidence interval:

{eq}\displaystyle (7.62\pm (1.669)(\frac{8.91}{\sqrt{65}}))\\(7.62\pm 1.84)\\P(5.78, 9.46) {/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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