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)

Explanation:

Confidence interval for population mean is given by :-

`\overline{x}\pm z_(\alpha/2)(\sigma)/(√(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)(11.8)/(√(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)