Answer:
Not enough evidence to reject Null hypothesis
Step-by-step explanation:
Solution:-
- A comparative study for bacterial growth in manual and electronic faucets is made.
- It is observed that there is a higher growth in electronic faucets due to slower flow rates, i.e electronic faucets are not thoroughly flushed; hence, giving more resident time for the scaled bacteria to grow.
- It is known that 15% of cultures from older faucets were tested positive for the Legionella bacteria.
- A study at John Hopkins was conducted on a sample n = 20 electronic faucets with the probability of bacteria growing in a faucet is 0.15.
- We will conduct a hypothesis for at-least half proportion of electronic faucets have cultured bacteria.
- State the hypothesis for the proportion of electronic faucets culturing Legionella bacteria:
Null Hypothesis: P = 0.15
Alternate hypothesis: P > 0.15
- To determine the test statistics for the study conducted at John hopkins. We had a sample size of n = 20, and the probability for a bacteria to grow in a faucet is 0.15.
- Denote random variable, X: The number of electronic faucets culturing Legionella bacteria.
- Since, the probability for a bacteria to grow in a faucet is independent for each new faucet. We will assume the RV " X " to follow binomial distribution with probability of success 0.15:
X ~ Bin ( 20 , 0.15 )
- We are to determine that at-least half of the sample is subjected to the said bacteria. This is the probability of P ( X ≥ 10 ).
- The pmf for a binomially distributed random variable X is given below:

Where,
p ( success ) = 0.15
p ( fail ) = 1 - p ( success ) = 1 - 0.15 = 0.85
- Use the pmf to determine the required test statistics:
![P ( X \geq 10 ) = 1 - P ( X \leq 9 )\\\\P ( X \geq 10 ) = 1 - [ (0.85)^2^0 + 20*(0.15)*(0.85)^1^9 + 20_C_2 (0.15)^2*(0.85)^1^8 +\\\\ 20_C_3 (0.15)^3*(0.85)^1^7 + 20_C_4 (0.15)^4*(0.85)^1^6 + 20_C_5 (0.15)^5*(0.85)^1^5+\\\\ 20_C_6 (0.15)^6*(0.85)^1^4 + 20_C_7 (0.15)^7*(0.85)^1^3 + 20_C_8 (0.15)^8*(0.85)^1^2 + \\\\ 20_C_9 (0.15)^9*(0.85)^1^1\\\\\\P ( X \geq 10 ) = 1 - [ 0.03875 + 0.13679 + 0.22933 + 0.24282 + 0.18212 + 0.10284 + \\\\ 0.04537 + 0.01601 + 0.00459 + 0.00108 ]\\\\](https://tex.z-dn.net/?f=P%20%28%20X%20%5Cgeq%2010%20%29%20%3D%201%20-%20P%20%28%20X%20%5Cleq%20%209%20%29%5C%5C%5C%5CP%20%28%20X%20%5Cgeq%2010%20%29%20%3D%201%20-%20%5B%20%280.85%29%5E2%5E0%20%2B%2020%2A%280.15%29%2A%280.85%29%5E1%5E9%20%2B%2020_C_2%20%280.15%29%5E2%2A%280.85%29%5E1%5E8%20%2B%5C%5C%5C%5C%2020_C_3%20%280.15%29%5E3%2A%280.85%29%5E1%5E7%20%2B%2020_C_4%20%280.15%29%5E4%2A%280.85%29%5E1%5E6%20%2B%2020_C_5%20%280.15%29%5E5%2A%280.85%29%5E1%5E5%2B%5C%5C%5C%5C%2020_C_6%20%280.15%29%5E6%2A%280.85%29%5E1%5E4%20%2B%2020_C_7%20%280.15%29%5E7%2A%280.85%29%5E1%5E3%20%2B%2020_C_8%20%280.15%29%5E8%2A%280.85%29%5E1%5E2%20%2B%20%5C%5C%5C%5C%2020_C_9%20%280.15%29%5E9%2A%280.85%29%5E1%5E1%5C%5C%5C%5C%5C%5CP%20%28%20X%20%5Cgeq%2010%20%29%20%3D%201%20-%20%5B%20%200.03875%20%2B%200.13679%20%2B%200.22933%20%2B%200.24282%20%2B%200.18212%20%2B%200.10284%20%2B%20%5C%5C%5C%5C%200.04537%20%2B%200.01601%20%2B%200.00459%20%2B%200.00108%20%5D%5C%5C%5C%5C)
![P ( X \geq 10 ) = 1 - [ 0.997 ] = 0.003](https://tex.z-dn.net/?f=P%20%28%20X%20%5Cgeq%2010%20%29%20%3D%201%20-%20%5B%200.997%20%5D%20%3D%200.003)
- The probability that 10 or more electronic faucets is found to have Legionella bacteria growing is 0.003
- The test proportion of 10 and more electronic faucets have culturing bacteria is p = 0.003.
- Assuming normality of the population, the Z-statistics would be:

- If we were to test the claim to 90% level of confidence:
significance level (α) = 1 - CI = 1 - 0.9 = 0.1
- The rejection region Z-critical is defined by a right-tail:
- Compare the test statistics with the rejection criteria defined by the Z-critical:
Z-test < Z-critical
-1.84 < 1.28
Conclusion:
There is not enough evidence that the probability of Legionella bacteria growing in electronic faucets is greater than 15%.