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The two nominal variables are related.
The following table summarises the data from a survey on the ownership of iPods among families with different levels of income.
Ownership C1 C2 C3
No 40 32 48
Yes 30 48 52
The first thing to do in order to determine if they are related or not is to state our null and alternative hypothesis
Null hypothesis
Alternative hypothesis
Using the Chi-square test statistics which can be expressed by using the formula
Ownership C1 C2 C3 Total
No 40 32 48 120
Yes 30 48 52 130
Total 70 80 100 250
The expected values are calculated as:
∴ Using the Chi-square test statistics, we have:
The degree of freedom df = ((r - 1) × (c - 1))
= (3 - 1) (2 -1 )
= 2 × 1
= 2
∴
Assuming the level of significance = 5%
The p-value of the Chi-square test statistics at df of 2 is:
=
Therefore, we can conclude that since the p-value (0.111) is greater than the level of significance (0.05), we fail to reject the null hypothesis.
Hence, the two nominal variables are related.
Learn more about Chi-square test statistics here:
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:
- 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%.