Respuesta :
Answer:
(32.2,34.7)
Step-by-step explanation:
Solution :
Given that,
\bar x = 33.4 hg
s = 6.4 hg
n = 177
Degrees of freedom = df = n - 1 = 177 - 1 = 176
At 99% confidence level the t is ,
α = 1 - 99% = 1 - 0.99 = 0.01
α / 2 = 0.01 / 2 = 0.005
tα /2,df = t0.005,176 = 2.604
Margin of error = E = tα/2,df * (s /√n)
= 2.604* ( 6.4/ √177)
= 1.25
The 95% confidence interval estimate of the population mean is,
\bar x - E < \mu < \bar x + E = 33.4 - 1.25 < \mu < 33.4 + 1.25
32.15 < \mu < 34.65
32.2 < \mu < 34.7
(32.2,34.7)
Question:
Here are summary statistics for randomly selected weights of newborn girls: n = 177, x = 28.9 hg, s = 6.7 hg. Construct a confidence interval estimate of the mean. Use a 98% confidence level. Are these results very different from the confidence interval 28.1 hg < μ < 30.7 hg with only 20 sample values, x' =29.4 hg, and s = 2.3 hg? What is the confidence interval for the population mean μ?
Answer:
The confidence interval for the population mean μ is 27.73 ≤ μ ≤ 30.0714
Step-by-step explanation:
The equation to identify the confidence interval for the mean is given by
[tex]x'-z_{\frac{\alpha }{2}} \frac{s}{\sqrt{n} } \leq \mu\leq x'+z_{\frac{\alpha }{2}} \frac{s}{\sqrt{n} }[/tex]
Where
x' = Sample mean = 28.9
s = Standard deviation = 6.7
n = Sample size = 177
[tex]z_{\frac{\alpha }{2}}[/tex] = Critical value = 2.326
Therefore we have
[tex]28.9-2.326\frac{6.7}{\sqrt{177} } \leq \mu\leq 28.9+2.326 \frac{6.7}{\sqrt{177} }[/tex]
27.73 ≤ μ ≤ 30.0714
T test we have
t = [tex]\frac{x'-\mu}{\frac{s}{\sqrt{n} } }[/tex]
=[tex]\frac{29.4-28.9}{\frac{2.3}{\sqrt{16} } }[/tex] = 0.8696 which is < 1
df = 15 as sample size = 15
Upper tail statistics lies between 0.3 and 0.1
