Question

Five measurements of the reaction time to a stimulus had a mean of 298 and a standard deviation of .024. Find a 90° confidence interval for the mean reaction time. An un contains an unknown proportion of red and white marbles. A random sample of 60 marbles selected with eplacement gave a result of 70% red. Find a 99% confidence interval for the proportion of red marbles in the urn. A sample of 40 was drawn from a population of 500. The sample mean was 400 and the standard deviation was 25. Find a 98% confidence interval for the mean.

Answer 1

i) Let $\mu$ denotes the mean reaction time.

Let ã denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, ã = 0.298, sa = 0.0006, s = 0.0240, n = 5 The margin of error is: 0.024 0.024 M E = - ✓n 9 *t 1-0.900 2 ,4 =- 5 * 10.050,4 = * 2.131847 = 0.022881 – 0.023 * 2.7 15 a 90 percent confidence interval population mean u is S = x + *t 1-0.900 = 0.298 + 0.024 * 0,050,4 = 0.298 - *t0.050.4 V5 = 0.298 +". 0.024 * 2.131847 15 = 0.298 + 0.022881 = (0.275119,0.320881) ~ (0.28,0.32)

ii) Let p denotes the true proportion of red marbles in the urn.

Let denotes the sample proportion and n denotes the sample size. Here, 0.7000 * (1 – 0.7000) P = 0.7000, n = 60, standard error = = 0.059161 60 Margin of error is Ô 0.700000 * (1 – 0.700000) ME= ME=> p = (1 - ) ...0 = 0.70 - (1 – 0.700), 2006 = þ* (1 – n * Z 1-0.99 = 0.700000 * (1 – 0.700000) 60 * 20.005 = 1 * 2.326348 = 0.137629 ** 60 a 99 percent confidence interval for the population proportion p is =p+/ 8*(1 – P) +21. 0.7000 + (1 - 0.7000) = 0.700 A * 20.005 60 -0.70 4. jaro 14 ore), asosan 0.7000 * (1 – 0.7000) * 2.575829 = 0.700 EM- 60 = 0.700 + 0.137629 = (0.562371, 0.837629) ~ (0.5624,0.8376)

iii)

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Let i denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, ū = 400.000, s2 = 625.0000, s = 25.0000, n = 40 The margin of error is: S 25.000 ME= 25.000 - 7*t 1-0.980 20 = *t0.010,39 = * 2.425841 = 9.588980 = 9.589 vn 140" 140 a 98 percent confidence interval population mean u is S = + *t 1-0.980 20 vn 2 ,39 25.000 = 400.000 + * 10.010,39 140 = 400.000 + 25.000 - * 2.425841 V40 = 400.000 = 9.588980 = (390.411020, 409.588980) – (390.41, 409.59)