Question

Here are summary statistics for randomly selected weights of newborn​ girls: nequals175​, x overbarequals30.1 ​hg, sequals7.1 hg. Construct a confidence interval estimate of the mean. Use a 98​% confidence level. Are these results very different from the confidence interval 29.0 hgless thanmuless than32.2 hg with only 20 sample​ values, x overbarequals30.6 ​hg, and sequals2.9 ​hg? What is the confidence interval for the population mean mu​?

Answer 1

sample mean, xbar = 30.1
sample standard deviation, s = 7.1
sample size, n = 175
degrees of freedom, df = n - 1 = 174

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.348

ME = tc * s/sqrt(n)
ME = 2.348 * 7.1/sqrt(175)
ME = 1.26

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (30.1 - 2.348 * 7.1/sqrt(175) , 30.1 + 2.348 * 7.1/sqrt(175))
CI = (28.84 , 31.36)

The reslts are different

Answer 2

 Identify the critical value (z)* for a 98% confidence level. For a normal distribution, this value is approximately 2.33.

1. Calculate the margin of error (E) 

2. using the formula:

   $$E=z^{\ast }\times \left(\frac{s}{\sqrt{n}}\right)=2.33\times \left(\frac{7.1}{\sqrt{175}}\right)\approx 1.25\text{hg}$$

3. Construct the confidence interval by adding and subtracting the margin of error from the sample mean:

   $$30.1\text{hg}\pm 1.25\text{hg}\Rightarrow (28.85\text{hg},31.35\text{hg})$$

Comparison with the Other Confidence Interval:

The provided interval (29.0 hg < μ < 32.2 hg) from a smaller sample (n = 20) has a wider range. This is expected because smaller samples have more variability, leading to less precision. Despite the difference in intervals, both include overlapping values (e.g., around 30–31 hg), so the results are not drastically different.

Final Answer:

The 98% confidence interval for the population mean μ is approximately 28.9 hg to 31.4 hg. This means we are 98% confident that the true average weight of newborn girls falls within this range. The difference from the smaller sample's interval is due to sample size and variability, but both estimates are reasonably consistent