identify the test statistics and p value. **Analysis of Strontium-90 in Baby Teeth from Two Cities**Listed in the data table are amounts of strontium-90 (in millibecquerels, or mBq, per gram of calcium) in a simple random sample of baby teeth obtained from residents in two cities. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Use a 0.10 significance level to test the claim that the mean amount of strontium-90 from city #1 residents is greater than the mean amount from city #2 residents.![Data Table Icon] Click the icon to view the data table of strontium-90 amounts.**Question:**What are the null and alternative hypotheses? Assume that population 1 consists of amounts from city #1 and population 2 consists of amounts from city #2.**Options:**- **A.** \( H_0: \mu_1 = \mu_2 \)- **B.** \( H_0: \mu_1 \ne \mu_2 \)- **C.** \( H_0: \mu_1 \le \mu_2 \)- **D.** \( H_0: \mu_1 \ge \mu_2 \)**Answer:**For this hypothesis test, the null and alternative hypotheses can be set up as follows:- **Null Hypothesis (\( H_0 \))**: \( \mu_1 \le \mu_2 \), which indicates that the mean amount of strontium-90 in residents of city #1 is less than or equal to that in city #2.- **Alternative Hypothesis (\( H_1 \))**: \( \mu_1 > \mu_2 \), which indicates that the mean amount of strontium-90 in residents of city #1 is greater than that in city #2.Thus, the correct answer from the provided options is **C.** ### Comparative Data of City #1 and City #2The table below presents a side-by-side comparison of numerical values associated with City #1 and City #2. The data is displayed in two columns, with each entry for City #1 corresponding to an equal number of entries for City #2.#### Data Table| City #1 | City #2 || -------- | -------- || 107 | 117 || 86 | 73 || 121 | 100 || 117 | 85 || 101 | 90 || 104 | 107 || 213 | 110 || 119 | 111 || 290 | 130 || 100 | 133 || 266 | 101 || 145 | 209 |This table can be used for a variety of analytical purposes such as comparing statistical measures (mean, median, mode), identifying trends, and performing data visualization exercises. **Explanation of Entries:**- Each row contains a pair of values, with the first number representing City #1 and the second representing City #2.- The data does not specify the exact nature of the values, leaving room for interpretation based on context such as population counts, temperature readings, sales numbers, etc.**Possible Analytical Exercises:**1. Calculate the mean, median, and mode for each city.2. Identify any significant differences or patterns between the two cities.3. Create visual representations such as bar graphs or line charts to illustrate the data comparisons.Use this data to enhance understanding of comparative analysis techniques and to practice interpreting statistical information.

The null and alternative hypotheses are: Here, µ1 is the mean amount from city #1, and µ2 is the…

Listed in the data table are amounts of strontium-90 (in millibecquerels, or mBq, per gram of calcium) in a simple random sample of baby teeth obtained from residents in two cities. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Use a 0.10 significance level to test the claim that the mean amount of strontium-90 from city #1 residents is greater than the mean amount from city #2 residents. E Click the icon to view the data table of strontium-90 amounts. What are the null and alternative hypotheses? Assume that population 1 consists of amounts from city #1 levels and population 2 consists of amounts from city #2.