146k views
1 vote
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

User Ischenkodv
by
8.1k points

1 Answer

3 votes

Answer:

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)\\\\

User Dylan Meivis
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories