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Math  /  Data & Statistics

QuestionThe 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. 5.20 5.72 4.62 4.80 5.02 4.60 4.74 5.19 4.87 4.76 4.56 5.70 Click the icon to view the table of critical t-values. (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 95%95 \% confidence that the population mean pH of rain water is between 4.73 and 5.24 . B. If repeated samples are taken, 95%95 \% of them will have a sample pH of rain water between □\square and □\square . C. There is a 95%95 \% probability that the true mean pH of rain water is between □\square and □\square . (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 99%99 \% confidence that the population mean pH of rain water is between □\square and □\square . B. There is a 99%99 \% probability that the true mean pH of rain water is between □\square and □\square . C. If repeated samples are taken, 99%99 \% of them will have a sample pH of rain water between □\square and □\square .

Studdy Solution

STEP 1

1. The sample size is n=12 n = 12 .
2. The data is normally distributed, and there are no outliers.
3. We will use the t-distribution to calculate confidence intervals due to the sample size being less than 30.

STEP 2

1. Calculate the sample mean and standard deviation.
2. Determine the critical t-value for 95% and 99% confidence intervals.
3. Calculate the margin of error for both confidence intervals.
4. Construct the confidence intervals.
5. Interpret the confidence intervals.

STEP 3

Calculate the sample mean (xˉ\bar{x}):
xˉ=5.20+5.72+4.62+4.80+5.02+4.60+4.74+5.19+4.87+4.76+4.56+5.7012 \bar{x} = \frac{5.20 + 5.72 + 4.62 + 4.80 + 5.02 + 4.60 + 4.74 + 5.19 + 4.87 + 4.76 + 4.56 + 5.70}{12}
xˉ=4.98 \bar{x} = 4.98
Calculate the sample standard deviation (ss):
s=∑(xi−xˉ)2n−1 s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
s≈0.42 s \approx 0.42

STEP 4

Determine the critical t-value for a 95% confidence interval with n−1=11 n-1 = 11 degrees of freedom. From the t-table, t0.025,11≈2.201 t_{0.025, 11} \approx 2.201 .
Determine the critical t-value for a 99% confidence interval with n−1=11 n-1 = 11 degrees of freedom. From the t-table, t0.005,11≈3.106 t_{0.005, 11} \approx 3.106 .

STEP 5

Calculate the margin of error for the 95% confidence interval:
E95=t0.025,11×sn E_{95} = t_{0.025, 11} \times \frac{s}{\sqrt{n}}
E95=2.201×0.4212 E_{95} = 2.201 \times \frac{0.42}{\sqrt{12}}
E95≈0.27 E_{95} \approx 0.27
Calculate the margin of error for the 99% confidence interval:
E99=t0.005,11×sn E_{99} = t_{0.005, 11} \times \frac{s}{\sqrt{n}}
E99=3.106×0.4212 E_{99} = 3.106 \times \frac{0.42}{\sqrt{12}}
E99≈0.38 E_{99} \approx 0.38

STEP 6

Construct the 95% confidence interval:
xˉ±E95=4.98±0.27 \bar{x} \pm E_{95} = 4.98 \pm 0.27
(4.71,5.25) (4.71, 5.25)
Construct the 99% confidence interval:
xˉ±E99=4.98±0.38 \bar{x} \pm E_{99} = 4.98 \pm 0.38
(4.60,5.36) (4.60, 5.36)

STEP 7

Interpret the 95% confidence interval:
A. There is 95% confidence that the population mean pH of rainwater is between 4.71 and 5.25.
Interpret the 99% confidence interval:
A. There is 99% confidence that the population mean pH of rainwater is between 4.60 and 5.36.
The 95% confidence interval is (4.71,5.25)(4.71, 5.25) and the 99% confidence interval is (4.60,5.36)(4.60, 5.36).

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