Question

1. a sample of the length in inches for newborns is given below. assume that lengths are normally distributed. find the 90% confidence interval of the mean length.

a. what is the point estimate for the population mean newborn length. (round answer to 2 decimal places)
b. which distribution should you use for this problem?
c. what must be true in order to construct a confidence interval in this situation?

• the population mean must be known
• the population must be approximately normal
• the sample size must be greater than 30
• the population standard deviation must be known

d. construct a 99% confidence interval for the population mean length of an infant. round upper and lower bounds to two decimal places.
e. what does it mean to be \99% confident\ in this problem?

• the researcher is 99% sure that the mean length of newborns is within this interval
• the researcher is 99% sure that the mean length for these 10 newborns is within this interval

Explanation:

Step1: Calculate the point estimate (sample mean)

The formula for the sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
Here, \(n = 10\) and \(x=\{22.2,17.3,15.7,18.1,16.6,22.2,15,21.9,19.5,21.9\}\)
\(\sum_{i=1}^{10}x_{i}=22.2 + 17.3+15.7+18.1+16.6+22.2+15+21.9+19.5+21.9 = 190.4\)
\(\bar{x}=\frac{190.4}{10}=19.04\)

Step2: Determine the distribution

Since the population standard deviation \(\sigma\) is unknown and the sample size \(n = 10<30\), we use the \(t -\)distribution.

Step3: Check the condition for confidence interval

For constructing a confidence interval for the population mean when \(\sigma\) is unknown and \(n<30\), the population must be approximately normal.

Step4: Construct the 99% confidence interval

The formula for the confidence interval is \(\bar{x}\pm t_{\alpha/2}\frac{s}{\sqrt{n}}\)
First, calculate the sample standard deviation \(s\)
\(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\(\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}=(22.2 - 19.04)^{2}+(17.3-19.04)^{2}+(15.7 - 19.04)^{2}+(18.1-19.04)^{2}+(16.6-19.04)^{2}+(22.2-19.04)^{2}+(15 - 19.04)^{2}+(21.9-19.04)^{2}+(19.5-19.04)^{2}+(21.9-19.04)^{2}\)
\(\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}=9.9856+3.0276 + 11.1556+0.8836+5.9536+9.9856+16.3216+8.1796+0.2116+8.1796=73.884\)
\(s=\sqrt{\frac{73.884}{9}}\approx2.86\)
For a 99% confidence interval, \(\alpha=1 - 0.99 = 0.01\), \(\alpha/2=0.005\), and \(n - 1=9\). From the \(t -\)table, \(t_{0.005,9}=3.250\)
The confidence interval is \(\bar{x}\pm t_{\alpha/2}\frac{s}{\sqrt{n}}=19.04\pm3.250\times\frac{2.86}{\sqrt{10}}\)
\(=19.04\pm3.250\times0.904\)
\(=19.04\pm2.94\)
The lower bound \(=19.04 - 2.94=16.10\)
The upper bound \(=19.04+2.94 = 21.98\)

Step5: Interpret 99% confidence

The 99% confidence means that the researcher is 99% sure that the mean length of newborns is within this interval.

Answer:

a. \(19.04\)
b. \(t -\)distribution
c. The population must be approximately normal
d. Lower bound \(=16.10\), Upper bound \(=21.98\)
e. The researcher is 99% sure that the mean length of newborns is within this interval