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.**Data:**5.05, 5.72, 5.24, 4.80 5.02, 4.58, 4.74, 5.19 5.29, 4.76, 4.56, 5.68 **(a) Determine a point estimate for the population mean.**A point estimate for the population mean is **5.05**. (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.)- O A. There is 95% confidence that the population mean pH of rainwater is between \_\_\_ and \_\_\_.- O B. There is a 95% probability that the true mean pH of rainwater is between \_\_\_ and \_\_\_.- O C. If repeated samples are taken, 95% of them will have a sample pH of rainwater between \_\_\_ and \_\_\_.(Note: The specific bounds for the confidence interval are not provided in the text and need to be calculated based on the data and appropriate formulae using the sample mean and standard deviation, along with the t-distribution critical values.)

Sample size is n=12 Sample data is x 5.05 5.72 5.24 4.8 5.02 4.58 4.74 5.19 5.29 4.76 4.56 5.68A point estimate of the population mean is the sample mean. Use Excel to obtain the sample mean. the steps are given below: Enter the sample data. Write the function "=AVERAGE()" and enter the whole range of the sample data as an argument of the function to obtain the sample mean. Therefore the sample mean is x¯=5.0525. A point estimate of the population mean is approximately 5.05 .Use Excel to obtain sample standard deviation. the steps are given below: Enter the sample data. Write the function "=STDEV.S()" and enter the whole range of the sample data as an argument of the function to obtain the sample standard deviation. Therefore the sample standard deviation is s=0.38889412 The degrees of freedom is df=n-1=12-1=11 Confidence level is 1-α=0.95 Significance level is α=0.05 Therefore, α2=0.025 Use Excel formula "=T.INV.2T(0.05,11)" to obtain the upper tail critical value of t-distribution with 11 degrees of freedom at 0.025 significance level. Therefore the critical value is tα2≅2.201 . The 95% confidence interval for population mean is C.I.=x¯±tα2sn≅5.0525±2.201×0.3888941212≈4.81 , 5.30 Interpretation: There is 95% confidence that the population mean PH of rain water is between 4.81 and 5.30. The correct option is option A.Confidence level is not a probability. Therefore, option B is incorrect. Confidence interval is for the population not for the sample. Therefore option C is incorrect.