Question

The mean monthly salary of a random sample of 20 college graduates under the age of 30 was found to be $1320 with a standard deviation of $677. Assume that the distribution of salaries for all college graduates under the age of 30 is normally distributed. Construct a 90% confidence interval for µ, the population mean of monthly salaries of all college graduates under the age of 30.

		Between $1071 and $1569
		Between $1408 and $4831
		Between $1058 and $1582
		Between $2858 and $3381
		Between $2741 and $3498 B) All other information remaining unchanged, which of the following would produce a narrower interval than the 90% confidence interval constructed above from Problem #4? A 95% confidence interval rather than a 90% confidence interval. A sample with a standard deviation of 725 instead of 677. A sample with a standard deviation of 1000 instead of 677. A sample of size 15 instead of 20. A sample of size 28 instead of 20

Answer 1

Answer (A)

One-Sample t-test Confidence Interval

The provided sample mean is Xˉ=1320 and the sample standard deviation is s=677. The size of the sample is n = 20 and the required confidence level is 90%. Degree of freedom The number of degrees of freedom are df = 20 - 1 = 19, and the significance level is α=0.1. Critical Value Based on the provided information, the critical t-value for α=0.1 and df=19 degrees of freedom is tc​=1.7291. Margin of Error Therefore, based on the information provided, the 90% confidence for the population mean μ is calculated as: Therefore, the 90% confidence interval for the population mean μ is 1058.2408<μ<1581.7592, which indicates that we are 90% confident that the true population proportion μ is contained by the interval

Answer (B)

A sample of size 28 instead of 20

The larger the sample size n, the more precise of an estimate can be obtained of a population parameter, via the use of confidence interval. Therefore the narrower is the width of the C.I

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