# Sample of two different models of cars were selected, and the actual speed for each car was determined when the speedometer registered 50 mph. The res

Sample of two different models of cars were selected, and the actual speed for each car was determined when the speedometer registered 50 mph. The resulting 95% confidence intervals for mean actual speed were (50.3, 52.7) and (49.8, 50.6) respectively. Assuming that the two sample standard deviations are equal, which confidence interval is based on the larger sample size? Explain your reasoning.
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Fatema Sutton
Step 1 The samples are selected for two different models of cars to check for the actual speed of the cars. The 95% confidence interval for mean actual speed was (50.3, 52.7) and (49.8,50.6). The general formula for finding the confidence interval for mean is, $\stackrel{―}{x}±$ (t critical value) $\frac{s}{\sqrt{n}}$ where s is the sample standard deviation. The width of the confidence interval is, 2(t critical value) $\frac{s}{\sqrt{n}}$ Step 2 Here, the two samples have equal standard deviations and hence, the narrower interval have larger sample size. The width of the first interval is $2.4\left(=52.7-50.3\right)$ and the second interval is $0.8\left(=50.6-49.8\right)$. Thus, the narrower interval is the second one. Thus, the confidence interval that is based on a larger sample is the second interval (49.8,50.6).