Confidence Interval (please show the work) The following sample of 16 measuremen

Question

Confidence Interval (please show the work)

The following sample of 16 measurements was selected from a population that is approximately normally distributed: S = {91 80 99 110 95 106 78 121 106 100 97 82 100 83 115 104} Construct a 80% confidence interval for the population mean. Interpret the meaning of this confidence interval for your STAT51 professor. The 95% confidence interval is: (91.19876, 104.6762). Explain why the 80% confidence interval is narrower than the 95% confidence interval. A random sample of 49 observations is drawn from a normal population with mean equal to 50 and sigma = 15. Find c such that P(x- lessthanorequalto c)= 89. Suppose T_10 is a Student's t distribution with 10 degrees of freedom. Find t_0 such that P(T_10 lessthanorequalto t_0) = .05.

Explanation / Answer

(8)

n = 16     

x-bar = 97.9375     

s = 12.646     

% = 80     

Standard Error, SE = s/n =    12.646/16 = 3.1615

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

t- score = 1.340605608     

Width of the confidence interval = t * SE =     1.34060560795885 * 3.1615 = 4.23832463

Lower Limit of the confidence interval = x-bar - width =      97.9375 - 4.23832462956189 = 93.69917537

Upper Limit of the confidence interval = x-bar + width =      97.9375 + 4.23832462956189 = 102.1758246

The 80% confidence interval is [93.70, 102.18]

(b)

We are 80% confident that the true (population) mean falls in the interval calculated above

(c)

80% confidence interval is narrower because the critical t- score for 80% confidence is 1.341, while for 95% confidence it is 2.131. The smaller t- score makes the interval narrower.

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