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13 January, 18:09

A university surveyed recent graduates of the English department for their starting salaries. Four hundred graduates returned the survey. The average salary was $25,000. The population standard deviation was $2,500. What is the 95% confidence interval for the mean salary of all graduates from the English department

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  1. 13 January, 19:41
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    Step-by-step explanation:

    We want to determine a 95% confidence interval for the mean salary of all graduates from the English department.

    Number of sample, n = 400

    Mean, u = $25,000

    Standard deviation, s = $2,500

    For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

    We will apply the formula

    Confidence interval

    = mean ± z * standard deviation/√n

    It becomes

    25000 ± 1.96 * 2500/√400

    = 25000 ± 1.96 * 125

    = 25000 ± 245

    The lower end of the confidence interval is 25000 - 245 = 24755

    The upper end of the confidence interval is 25000 + 245 = 25245

    Therefore, with 95% confidence interval, the mean salary of all graduates from the English department is between $24755 and $25245
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