Test the CDC's claim that at least 64% of adults wash their hands regularly
The CDC states that the proportion of adults that wash their hands regularly is at least 64%. You would like to test this claim at the .05 level of significance, so you randomly select 240 adults and find that 138 wash their hands regularly. Conduct a hypothesis test for The CDCs claim (Round the sample proportion to 4 decimal places when using in the test statistic) Chris OByrne 2024
A. What will be the alternative hypothesis? (just need to state 1-tail -Lower, 1-tail-Upper or 2-tail - just put one of those 3 answers in the body of the discussion board).
B. The value of the test statistic and which test statistic you used (example: z = 1.23 or t = 1.234) and make sure to round correctly depending on which test statistic you will be using.
C. What is the probability of the test statistic for this specific problem? (take into consideration if this is 1-tail or 2-tail test)
D. What conclusion will you come to? (just need to state Accept Ho or Reject Ho and Accept Ha just put one of those 2 answers in the body of the discussion board).
E. Using the critical value approach, at what critical value or values will you start rejecting the null hypothesis? (make sure you are very careful and detailed with your answer, pay attention to signs and values)
To test the CDC's claim that at least 64% of adults wash their hands regularly, we will conduct a hypothesis test using the information provided.
A. Alternative Hypothesis
Since we want to test if the proportion of adults who wash their hands regularly is less than 64%, our alternative hypothesis will be:
- 1-tail -Lower
B. Test Statistic Calculation
1. Sample Proportion (p̂):
[
p̂ = \frac{x}{n} = \frac{138}{240} = 0.5750
]
(rounded to four decimal places)
2. Null Hypothesis (H₀):
[
H₀: p \geq 0.64
]
3. Alternative Hypothesis (H₁):
[
H₁: p < 0.64
]
4. Population Proportion (p₀):
( p₀ = 0.64 )
5. Standard Error (SE):
[
SE = \sqrt{\frac{p₀(1 - p₀)}{n}} = \sqrt{\frac{0.64 \times (1 - 0.64)}{240}} = \sqrt{\frac{0.64 \times 0.36}{240}} = \sqrt{\frac{0.2304}{240}} \approx 0.0470
]
6. Test Statistic (z):
[
z = \frac{p̂ - p₀}{SE} = \frac{0.5750 - 0.64}{0.0470} \approx \frac{-0.0650}{0.0470} \approx -1.3830
]
(rounded to four decimal places)
C. Probability of the Test Statistic
To find the probability associated with our test statistic, we will use the z-table to find the p-value for ( z = -1.3830 ).
- Looking up ( z = -1.38 ) in a z-table gives a cumulative probability of approximately 0.0836.
Since this is a one-tailed test, the probability (p-value) associated with the test statistic is:
- p-value ≈ 0.0836
D. Conclusion
Given the significance level ( \alpha = 0.05 ):
- Since ( p-value (0.0836) > \alpha (0.05) ), we do not have enough evidence to reject the null hypothesis.
Therefore, we conclude:
- Accept H₀
E. Critical Value Approach
For a one-tailed test at the 0.05 significance level, we need to find the critical z-value:
- The critical value for a one-tailed test at ( \alpha = 0.05 ) is approximately ( z_{critical} = -1.645).
Decision Rule:
- If ( z < -1.645 ), reject ( H₀ ).
- If ( z ≥ -1.645 ), accept ( H₀ ).
In this case, since ( z = -1.3830 > -1.645 ):
- We do not reject the null hypothesis.
Summary
- A: 1-tail -Lower
- B: z = -1.3830
- C: p-value ≈ 0.0836
- D: Accept H₀
- E: Critical value is -1.645
This analysis supports the claim that at least 64% of adults wash their hands regularly, based on the sample data provided.