Question

The following data represent the pH of rain for a random sample of 12 rain dates, A normal probability plot suggests the data could come from a population that is normaly distributed. A boxplot indicates there are no outliers. Complete parts (a) through (d) below. 5.20 5.02 4.87 5.72 4.57 4.76 4.89 474 4.56 4.80 5.19 5.30 (a) Determine a point estimate for the population mean. A point estimate for the population mean is Round to two decimal places as needed.) Construct and interpret a 95% confidence interval for the mean pH of rainwater. Select the correct choice below and n11 in the answer boxes to complete your co (Use ascending order. Round to two decimal places as needed.) A. There is a 95% probability that the true mean pH of rain water is between and OB There is 95% confidence that the population mean pH of rain water!s between land If repeated samples are taken, 95% of them will have a sample pH of rain water between C. and c) Construct and interpret a gg% confidence interval for the mean pH of rainwater. Select the correct choice below and fil in the answer boxes to complete your choice. (Use ascending order. Round to two decimal places as needed.) A. There is 99% contdence that the population mean pH of rain water is between and B. If repeated samples are taken, 99% or them wil have a sample pH of rain water between and ° C. There is a 99% probability that the true mean pH of rain water is between-and (d) What happens to the interval as the level of confidence is changed? Explain why this is a logical result As the level of confidence increases, the width of the interval This makes sense since

Answer 1

(a) The computati on table 5.2 5.02 4.87 5.72 4.57 4.76 4.89 4.74 4.56 4.8 5.19 5.3 0.231667 0.051667 -0.09833 0.751667 -0.39833 -0.20833 -0.07833 -0.22833 -0.40833 -0.16833 0.221667 0.331667 0.05367 0.002669 0.009669 0.565003 0.158669 0.043403 0.006136 0.052136 0.166736 0.028336 0.049136 0.110003 12 12 2X, 59.62 (x-X2 1.245567 The point estim ate of the population mean is il 12 = (59.62) 4.968333 X = 4.9683331 b) The point estimate of the standard deviati on is 12 12-1 1.245567 V0.113233 0.336502 ơ~10. 336502 Therefore, the standard deviation is approximately Ơ~ 10.336502

b) The sample size is n 12. The sample mean isX 4.968333 The sample standard deviation is s 0.336502. The degrees of freedom is df n-1 - 12-1 The level of significance is a-0.05 The table values are cal culated by using the Excel2010 function, =tiny(probability, degrees of freedom ti(0.05,11) -2.200985 - 2.200985 The formula for calculating the 95% confidence interval for the sample mean is 0.336502:s (0, 202)sus4 968333+2 200985(03 202) 4.968333- 2.200985 4968333- 3.464102 0.740635 0.740635 3.464102 4.968333 (0.213803) SHS4.968333+(0.213803) 4.75453SHS5.182136 Interpretati on: We can be 95% confident that the population mean PH of rain water is between 4.7545 and 5.1821

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c) The sample size is n 12. The sample mean is X= 4.968333 The sample standard deviation is s= 0.336502. The degrees of freedom is df n-1 =12-1 The level of significance is a 0.01 The table values are cal culated by using the Excel2010 function, tinv(probability, deg rees of jreedom) -tin(0.01,11) 3.105807 - 3105807 The formula for calculating the 99% confidence interval for the sample mean is 0.33650249 104100 0.336502 4.968333-3.105807 S4.968333+3.105807 4.968333+1045109 3464 102 μ 3.464102 4.968333 (0.301697)S us4.968333 (0.301697) 4.666636 SuS5.27003 Interpretati on: We can be 99% confident that the population mean PH of rain water is between 4.6666 and 5.2700 (d) The parts (a) and (b) enables that, As the level of confi dence increases, the width of the interval increases. This makes sense, since the width of the confidence interval increases as the level of significance increases