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27 July, 17:21

In an examination of purchasing patterns of shoppers, a sample of 20 shoppers revealed that they spent, on average, $54 per hour of shopping. Based on previous years, the population standard deviation is thought to be $21 per hour of shopping. Assuming that the amount spent per hour of shopping is normally distributed, find a 90% confidence interval for the mean amount.

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  1. 27 July, 20:39
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    Confidence interval for the mean amount = 54+1.645*21/sqrt (16) = (62.64, 45.36)

    Explanation:

    confidence interval = mean + z*, where z * is the upper (1-C) / 2 critical value for the standard normal distribution.

    z score for 90% confidence interval = 1.645

    confidence interval for the mean amount = 54+1.645*21/sqrt (16) = (62.64, 45.36)
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