Question

You measure 44 textbooks' weights, and find they have a mean weight of 51 ounces. Assume the population standard deviation is 11.8 ounces. Based on this, construct a 95% confidence interval for the true population mean textbook weight. Give your answers as decimals, to two places

Answer 1

Answer: (47.51, 54.49)

Step-by-step explanation:

Confidence interval for population mean is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

, where n= sample size .

$\sigma$ = population standard deviation.

$\overline{x}$ = sample mean

$z_{\alpha/2}$ = Two -tailed z-value for ${\alpha$ (significance level)

As per given , we have

$\sigma=11.8\text{ ounces}$

$\overline{x}=51 \text{ ounces}$

n= 44

Significance level for 95% confidence = $\alpha=1-0.95=0.05$

Using z-value table ,

Two-tailed Critical z-value : $z_{\alpha/2}=z_{0.025}=1.96$

Now, the 95% confidence interval for the true population mean textbook weight will be :-

$51\pm (1.96)\dfrac{11.8}{\sqrt{44}}\\\\=51\pm(1.96)(1.7789)\\\\=51\pm3.486644\approx51\pm3.49\\\\=(51-3.49,\ 51+3.49)\\\\=(47.51,\ 54.49)$

Hence, the 95% confidence interval for the true population mean textbook weight. :  (47.51, 54.49)