Question

A random sample of n measurements was selected from a population with unknown mean mu and standard deviation sigmaequals50 for each of the situations in parts a through d. Calculate a 95​% confidence interval for mu for each of these situations. a. nequals40​, x overbarequals42 b. nequals300​, x overbarequals123 c. nequals155​, x overbarequals20 d. nequals155​, x overbarequals3.14 e. Is the assumption that the underlying population of measurements is normally distributed necessary to ensure the validity of the confidence intervals in parts a through​ d? Explain.

Answer 1

Answer:

a) (26.50;57.50)

b) (117.34;128.66)

c) (12.13;27.87)

d) (-4.73;11.01)

e) No. Since the sample sizes are large (n ≥ 30), the central limit theorem  guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval is given by this formula:

$\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$   (1)

And for a 95% of confidence the significance is given by $\alpha=1-0.95=0.05$, and $\frac{\alpha}{2}=0.025$. Since we know the population standard deviation we can calculate the critical value $z_{0.025}= \pm 1.96$

Part a

$n=40,\bar X=42,\sigma=50$

If we use the formula (1) and we replace the values we got:

$42 - 1.96 \frac{50}{\sqrt{40}}=26.50$  

$42 + 1.96 \frac{50}{\sqrt{40}}=57.50$  

The 95% confidence interval is given by (26.50;57.50)

Part b

$n=300,\bar X=123,\sigma=50$

If we use the formula (1) and we replace the values we got:

$123 - 1.96 \frac{50}{\sqrt{300}}=117.34$  

$123 + 1.96 \frac{50}{\sqrt{300}}=128.66$  

The 95% confidence interval is given by (117.34;128.66)

Part c

$n=155,\bar X=20,\sigma=50$

If we use the formula (1) and we replace the values we got:

$20 - 1.96 \frac{50}{\sqrt{155}}=12.13$  

$20 + 1.96 \frac{50}{\sqrt{155}}=27.87$  

The 95% confidence interval is given by (12.13;27.87)

Part d

$n=155,\bar X=3.14,\sigma=50$

If we use the formula (1) and we replace the values we got:

$3.14 - 1.96 \frac{50}{\sqrt{155}}=-4.73$  

$3.14 + 1.96 \frac{50}{\sqrt{155}}=11.01$  

The 95% confidence interval is given by (-4.73;11.01)

Part e

No. Since the sample sizes are large (n ≥ 30), the central limit theorem  guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid