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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 normally distributed. A boxplot indicates there are no outliers. Complete parts (a) through (d) below. 5.05 5.72 4.62 4.80 5.02 4.57 4.74 5.19 4.61 4.76 4.56 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.) (b) Construct and interpret a 95% confidence interval for the mean pH of rainwater. Select the correct choice below and fill in the answer boxes to complete your choice. (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 ◻ B. If repeated samples are taken, 95% of them will have a sample pH of rain water between ◻ and ◻ C. There is 95% confidence that the population mean pH of rain water is between ◻ and ◻ (c) Construct and interpret a 99% confidence interval for the mean pH of rainwater. Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to two decimal places as needed.) A. There is a 99% probability that the true mean pH of rain water is between ◻ and ◻ . B. There is 99% confidence that the population mean pH of rain water is between ◻ and ◻ C. If repeated samples are taken, 99% of them will have a sample pH of rain water between ◻ and ◻

The following data represent the pH \mathrm{pH} of rain for a random sample of 1212 rain dates. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts (a) through (d) below.\newline\begin{tabular}{llll}\newline55.0505 & 55.7272 & 44.6262 & 44.8080 \\\newline55.0202 & 44.5757 & 44.7474 & 55.1919 \\\newline44.6161 & 44.7676 & 44.5656 & 55.3030\newline\end{tabular}\newline(a) Determine a point estimate for the population mean.\newlineA point estimate for the population mean is □ \square (Round to two decimal places as needed.)\newline(b) Construct and interpret a 9595\% confidence interval for the mean pH \mathrm{pH} of rainwater. Select the correct choice below and fill in the answer boxes to complete your choice.\newline(Use ascending order. Round to two decimal places as needed.)\newlineA. There is a 95% 95 \% probability that the true mean pH \mathrm{pH} of rain water is between □ \square and □ \square \newlineB. If repeated samples are taken, 95% 95 \% of them will have a sample pH \mathrm{pH} of rain water between □ \square and □ \square \newlineC. There is 95% 95 \% confidence that the population mean pH \mathrm{pH} of rain water is between □ \square and □ \square \newline(c) Construct and interpret a 9999\% confidence interval for the mean pH \mathrm{pH} of rainwater. Select the correct choice below and fill in the answer boxes to complete your choice.\newline(Use ascending order. Round to two decimal places as needed.)\newlineA. There is a □ \square 66 probability that the true mean pH \mathrm{pH} of rain water is between □ \square and □ \square .\newlineB. There is □ \square 66 confidence that the population mean pH \mathrm{pH} of rain water is between □ \square and □ \square \newlineC. If repeated samples are taken, □ \square 66 of them will have a sample pH \mathrm{pH} of rain water between □ \square and □ \square

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Q. The following data represent the pH \mathrm{pH} of rain for a random sample of 1212 rain dates. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts (a) through (d) below.\newline\begin{tabular}{llll}\newline55.0505 & 55.7272 & 44.6262 & 44.8080 \\\newline55.0202 & 44.5757 & 44.7474 & 55.1919 \\\newline44.6161 & 44.7676 & 44.5656 & 55.3030\newline\end{tabular}\newline(a) Determine a point estimate for the population mean.\newlineA point estimate for the population mean is □ \square (Round to two decimal places as needed.)\newline(b) Construct and interpret a 9595\% confidence interval for the mean pH \mathrm{pH} of rainwater. Select the correct choice below and fill in the answer boxes to complete your choice.\newline(Use ascending order. Round to two decimal places as needed.)\newlineA. There is a 95% 95 \% probability that the true mean pH \mathrm{pH} of rain water is between □ \square and □ \square \newlineB. If repeated samples are taken, 95% 95 \% of them will have a sample pH \mathrm{pH} of rain water between □ \square and □ \square \newlineC. There is 95% 95 \% confidence that the population mean pH \mathrm{pH} of rain water is between □ \square and □ \square \newline(c) Construct and interpret a 9999\% confidence interval for the mean pH \mathrm{pH} of rainwater. Select the correct choice below and fill in the answer boxes to complete your choice.\newline(Use ascending order. Round to two decimal places as needed.)\newlineA. There is a □ \square 66 probability that the true mean pH \mathrm{pH} of rain water is between □ \square and □ \square .\newlineB. There is □ \square 66 confidence that the population mean pH \mathrm{pH} of rain water is between □ \square and □ \square \newlineC. If repeated samples are taken, □ \square 66 of them will have a sample pH \mathrm{pH} of rain water between □ \square and □ \square
  1. Calculate sum of pH values: Calculate the sum of all pH values to find the total sum for the sample.\newlineSum = 5.05+5.72+4.62+4.80+5.02+4.57+4.74+5.19+4.61+4.76+4.56+5.30=58.945.05 + 5.72 + 4.62 + 4.80 + 5.02 + 4.57 + 4.74 + 5.19 + 4.61 + 4.76 + 4.56 + 5.30 = 58.94
  2. Determine point estimate: Determine the point estimate for the population mean by dividing the total sum by the number of observations.\newlinePoint estimate = 58.9412=4.91\frac{58.94}{12} = 4.91
  3. Calculate sample standard deviation: Calculate the sample standard deviation. First, find the squared deviations from the mean for each data point.\newlineSquared deviations = (5.05−4.91)2+(5.72−4.91)2+(4.62−4.91)2+(4.80−4.91)2+(5.02−4.91)2+(4.57−4.91)2+(4.74−4.91)2+(5.19−4.91)2+(4.61−4.91)2+(4.76−4.91)2+(4.56−4.91)2+(5.30−4.91)2=1.2068(5.05 - 4.91)^2 + (5.72 - 4.91)^2 + (4.62 - 4.91)^2 + (4.80 - 4.91)^2 + (5.02 - 4.91)^2 + (4.57 - 4.91)^2 + (4.74 - 4.91)^2 + (5.19 - 4.91)^2 + (4.61 - 4.91)^2 + (4.76 - 4.91)^2 + (4.56 - 4.91)^2 + (5.30 - 4.91)^2 = 1.2068
  4. Finish calculating standard deviation: Finish calculating the sample standard deviation.\newlineStandard deviation = 1.2068(12−1)=0.33\sqrt{\frac{1.2068}{(12 - 1)}} = 0.33
  5. Construct 9595% confidence interval: Construct a 9595% confidence interval for the mean pH of rainwater using the t-distribution (t-value approximately 2.2012.201 for df=11df = 11).\newlineMargin of error = 2.201×(0.33/12)=0.212.201 \times (0.33 / \sqrt{12}) = 0.21
  6. Calculate lower and upper bounds: Calculate the lower and upper bounds of the 9595% confidence interval.\newlineLower bound = 4.91−0.21=4.704.91 - 0.21 = 4.70; Upper bound = 4.91+0.21=5.124.91 + 0.21 = 5.12
  7. Construct 9999% confidence interval: Construct a 9999% confidence interval for the mean pH of rainwater using the t-distribution (t-value approximately 3.1063.106 for df=11df = 11).\newlineMargin of error = 3.106×(0.33/12)=0.303.106 \times (0.33 / \sqrt{12}) = 0.30
  8. Calculate lower and upper bounds: Calculate the lower and upper bounds of the 99%99\% confidence interval.\newlineLower bound = 4.91−0.30=4.614.91 - 0.30 = 4.61; Upper bound = 4.91+0.30=5.214.91 + 0.30 = 5.21

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