x = 21.29 s = 1.80 The critical value for a 99% confidence interval based on 6 d

Question

x = 21.29 s = 1.80 The critical value for a 99% confidence interval based on 6 df is 3.71. The interval is x bar plusminus (t critical value) (s/squareroot nm) = 21.29 plusminus (3.71) (1.80/squareroot 7) = 21.29 plusminus 2.52 = (18.77, 23.81) A statistical software package could also have been used to compute the 99% confidence interval. The following is output from SPSS. The slight discrepancy between the hand-calculated interval and the one reported by SPSS occurs because SPSS uses more decimal accuracy in x bar, sigma, and t critical values. with 99% confidence, we estimate the population mean number of charitable responses (out of 36 trials) to be between 18, 77 and 23, 81. Remember that the 99% confidence level implies that if the same formula is used to calculate intervals for sample after sample randomly selected from the population of chimps, in the long run 99% of these intervals will capture the value of mu between the lower and upper confidence limits. Notice that based on this interval, we would conclude that, on average, chimps choose the charitable option more than half the time (18 out of 36 trials). The Newsday headline "Chimps Aren't Charitable" was based on additional data from the study indicating that chimps' charitable behavior was no different when there was another chimp in the adjacent cage than when the adjacent cage was empty. How did we obtain the value of df? It would shrink to sample mean. df = n - 1, where n is the size of the sample. df = n - 1, where n is the size of the population. df is always equal to 6. Remember the general properties of t distribution. Without doing any calculations, how do you think the interval would change if s were 2.50 rather than 1.80? It would shift to the right, the length would not change. It would widen, not shifting anywhere. It would shrink, not shifting anywhere. It would shift to the left, the length would not change.

Explanation / Answer

First part - For a t distribution, the degrees of freedom is equal to sample size - 1

So option B is correct i.e. df= n - 1 where n is the size of the sample.

Second part - If the value of standard deviation increases from 1.8 to 2.5, then Margin of error will increase beacuase Margin of error = Critical value x ( Standard deviation of the statistic/sqrt(n) )

So as the value of Margin of error increases the confidence interval will widen as the value of confidence interval is : sample statistic + margin of error. So option B is correct i.e it would widen, not shifting anywhere.

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