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

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

Explanation:

Given:

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

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

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

`E = Z_{(\alpha)/(2) * ( (\sigma)/(√(n)))`

`= 1.645 * ((4)/(√(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)\\\\`