Question

The lengths of text messages are normally distributed with a population standard deviation of 4 characters and an unknown population mean. If a random sample of 26 text messages is taken and results in a sample mean of 28 characters, find a 98% confidence interval for the population mean. Round your answers to two decimal places. z0.10 z0.05 z0.04 z0.025 z0.01 z0.005 1.282 1.645 1.751 1.960 2.326 2.576

Answer 1

Answer:

The answer is "$(25.32, 30.68)$"

Step-by-step explanation:

Given:

$\bar x = 28\\\\ \sigma = 4\\\\n =26$

When $98\%$ confidence level so, the z:

$\alpha = 1 - 98\% = 1 - 0.98 = 0.2\\\\\frac{\alpha}{2} =\frac{0.02}{2} = 0.01\\\\Z_{\frac{\alpha}{2}} = Z_{0.01} = 1.645\ \ \ ( Using \ z \ table )$

$E = Z_{\frac{\alpha}{2} \times ( \frac{\sigma}{\sqrt{n}})$  

$= 1.645 \times (\frac{4}{\sqrt{26}})\\\\=2.68$

When  98% confidence interval estimate of the population mean  is,

$\bar x - E < \mu < \bar x + E\\\\28 - 2.68 < \mu < 28 + 2.68\\\\25.32 < \mu < 30.68\\\\(25.32, 30.68)$