Find a 90% confidence interval for the mean number of days, mu, of diverted medi

Question

Find a 90% confidence interval for the mean number of days, mu, of diverted medical marijuana use in the last 6 months of all adolescents in substance abuse treatment. The 90 % confidence interval for mu is from days to days. C. Draw a graph to display both confidence intervals. D. Which confidence interval yields a more accurate estimate of mu? The % confidence interval is a more accurate estimate, because it is than the % confidence interval. Sec 8.3 A variable has a mean of 100 and a standard deviation of 16. Sixteen observations of this variable have mean of 112 and a sample standard deviation of 24. Determine the observed value of the A. Standardized version of x bar, z = B. Studentized version of x bar, t = For a t-curve with df = 17, use a t-distribution table to find the t-value t_0.025 Use the one-mean t-interval procedure with the sample mean, sample size, sample standard deviation, and confidence level given below. x bar = 21, n = 16, s = 5, confidence level = 98% A. Find a confidence interval for the mean of the population from which the sample was drawn. (Round to three decimal places as needed) The 98% confidence interval about mu is to B. Obtain the margin of error by taking half the length of the confidence interval. (Round to three decimal places as needed) C. Obtain the margin of error by taking the formula t_alpha/2 middot s/squareroot pi (Round to three decimal places as needed)

Explanation / Answer

Solution10:

t critical for 0.025 level of significance and 17 df is

=2.110

Solution11:

n=16<30 use t test

alpha=1-98%=1-0.98=0.02

alpha/2=0.02/2=0.01

degrees of freedom=n-1=16-1=15

t critical=2.602

confidence interval for mean is given by formula

=21+-2.602(5/sqrt(16))

=21-3.2525<mu<21+3.2525

=17.748<mu<24.253

lower limit=17.748

upper limit=24.253

margin of error=3.253

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