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Confidence Intervals for Means - Remember: Sampling Distribution of Means

Learn how to construct confidence intervals for means and understand the differences between means and proportions. Find out how to calculate confidence intervals using the z-distribution and t-distribution. Also, discover how to determine the minimum required sample size.

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Confidence Intervals for Means - Remember: Sampling Distribution of Means

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  1. Confidence Intervals Means

  2. Remember: In sampling distribution of means….. • Approximately normal if • n is large ( n 30) – CLT • The population is normal

  3. So how will I find a confidence intervals for a mean? How is it different from a proportion?

  4. Formula for a z-confidence interval:

  5. Suppose you work for a consumer advocate agency & want to find the mean repair cost of a washing machine. You randomly select 40 repair costs and find the mean to be $100. The standard deviation is $17.50. construct a 95% interval for the population mean.

  6. Find the minimum required sample size if you want to be 95% confident that the sample mean is within 2 units of the population mean if the population standard deviation is 4.8.

  7. But what if the sample size is small or the population standard deviation is unknown?

  8. We use the t-distribution if the population standard deviation is unknown. • It’s bell shaped – centered at 0. • Each t-distribution is more spread out than the z-distribution (normal). • As the sample size increases, the spread decreases and actually approaches the normal distribution. • Based on degrees of freedom (df = n-1)

  9. Degrees of freedom: (n-1) • Suppose A + B + C + D = 18 • Thus there are 3 degrees of freedom Free to be anything

  10. The t- distribution compared to the normal distribution • http://www.nku.edu/~longa/stats/taryk/TDist.html • In normal sampling but with t-distribution

  11. Reading the t-chart

  12. The Raman Arches is an Italian restaurant. The manager wants to estimate the average amount a customer spends on lunch. A random sample of 115 customers’ lunch tabs gave a mean of $9.74 with a standard deviation of $2.93. Find and interpret a 90% confidence interval for the average amount spent on lunch by all customers.

  13. A sample of 20 students had a test average of 100 with a standard deviation of 4.2 points. Find and interpret a 90% confidence interval for the average test score for all students on this test.

  14. A random sample of large tents listed in Consumer Reports: Special Outdoor Issue gave the following prices. Find a 90% confidence interval for the mean price of all such tents.

  15. Homework - Worksheet

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