The Amount Of Water Consumed Each Day By A Healthy Adult
The Amount Of Water Consumed Each Day By A Healthy Adult Follows A Nor
The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.38 liters. A sample of 10 adults after a campaign shows the following consumption in liters: 1.48, 1.50, 1.42, 1.40, 1.56, 1.80, 1.62, 1.84, 1.42, 1.39. At the 0.05 significance level, we are to determine whether water consumption has increased, which involves conducting a hypothesis test for the mean.
Hypotheses Formulation
The null hypothesis (H₀) assumes that there has been no increase in the average water consumption, remaining at the original mean of 1.38 liters. The alternative hypothesis (H₁) posits that the mean has increased following the campaign, making it a one-sided (upper-tail) test. Specifically:
- Null hypothesis: H₀: μ = 1.38
- Alternative hypothesis: H₁: μ > 1.38
These are expressed with the notations:
```plaintext
H₀: μ = 1.38
H₁: μ > 1.38
```
Decision Rule at the 0.05 Significance Level
Given the sample size (n = 10), we will perform a t-test for the mean, as the population standard deviation is unknown. The decision rule involves computing the critical t-value (t-critical) for a right-tailed test with 9 degrees of freedom (n - 1) at the 0.05 significance level. Using a t-distribution table or calculator, the critical value is approximately:
- t-critical (df = 9, α = 0.05): 1.833
If the calculated t-statistic exceeds this critical value, we reject the null hypothesis.
Calculation of the Test Statistic
First, compute the sample mean (\(\bar{x}\)):
\[
\bar{x} = \frac{1.48 + 1.50 + 1.42 + 1.40 + 1.56 + 1.80 + 1.62 + 1.84 + 1.42 + 1.39}{10}
\]
\[
\bar{x} = \frac{14.83}{10} = 1.483
\]
Next, calculate the sample standard deviation (s):
Calculate each squared deviation:
\[
(1.48 - 1.483)^2 = 0.00009
\]
\[
(1.50 - 1.483)^2 = 0.000289
\]
\[
(1.42 - 1.483)^2 = 0.003769
\]
\[
(1.40 - 1.483)^2 = 0.006889
\]
\[
(1.56 - 1.483)^2 = 0.006289
\]
\[
(1.80 - 1.483)^2 = 0.100489
\]
\[
(1.62 - 1.483)^2 = 0.017489
\]
\[
(1.84 - 1.483)^2 = 0.128449
\]
\[
(1.42 - 1.483)^2 = 0.003769
\]
\[
(1.39 - 1.483)^2 = 0.008409
\]
Sum of squared deviations:
\[
0.00009 + 0.000289 + 0.003769 + 0.006889 + 0.006289 + 0.100489 + 0.017489 + 0.128449 + 0.003769 + 0.008409 = 0.275
\]
Sample variance:
\[
s^2 = \frac{0.275}{n - 1} = \frac{0.275}{9} \approx 0.030556
\]
Standard deviation:
\[
s = \sqrt{0.030556} \approx 0.175
\]
Compute the t-statistic:
\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{1.483 - 1.38}{0.175 / \sqrt{10}} = \frac{0.103}{0.175 / 3.162} = \frac{0.103}{0.0554} \approx 1.86
\]
Interpretation and p-value
The calculated t-value is approximately 1.86, which is slightly above the critical value of 1.833. Since it exceeds the critical value, we reject the null hypothesis at the 0.05 significance level, indicating evidence that water consumption has increased after the campaign.
The p-value corresponding to t = 1.86 with 9 degrees of freedom can be looked up using a t-distribution calculator or software. The p-value for a one-sided test:
- p-value ≈ 0.048
This indicates that there is about a 4.8% chance of observing such data if the true mean had remained at 1.38 liters.
Conclusion
Based on the analysis, since the test statistic exceeds the critical value and the p-value is less than 0.05, we reject the null hypothesis. There is statistically significant evidence at the 5% level to conclude that the campaign has led to an increase in the average water consumption among adults.
Paper For Above instruction
The study of water consumption patterns among healthy adults is crucial for understanding hydration behaviors and informing public health strategies. This research focuses on evaluating whether a recent campaign has effectively increased the average daily water intake among adults, utilizing statistical hypothesis testing to assess the validity of observed changes.
Prior to the campaign, the population's mean water intake was established at 1.38 liters per day, based on historical data suggesting normal distribution characteristics. To determine if the campaign influenced water consumption, a sample of 10 adults was surveyed, yielding individual daily intake measurements: 1.48, 1.50, 1.42, 1.40, 1.56, 1.80, 1.62, 1.84, 1.42, and 1.39 liters. The primary goal was to test whether this sample provides evidence of an increased mean compared to the pre-campaign mean.
The null hypothesis (H₀) states that the mean water intake remains unchanged at 1.38 liters, while the alternative hypothesis (H₁) asserts that the mean has increased. Formally, H₀: μ = 1.38 versus H₁: μ > 1.38. This constitutes a one-sided t-test, appropriate given the small sample size and unknown population variance.
The decision rule involves computing the t-statistic for the sample mean and comparing it to the critical t-value at the 0.05 significance level with 9 degrees of freedom (n - 1). The critical t-value for a right-tailed test at this significance level is approximately 1.833. If the calculated t-statistic exceeds this threshold, the null hypothesis is rejected, indicating statistically significant evidence of increased water consumption.
Calculations begin with the sample mean, which, summing the observed values, yields 1.483 liters. Calculating the sample standard deviation involves evaluating deviations from the mean and their squared differences, resulting in an estimated standard deviation of approximately 0.175 liters. Using these figures, the t-statistic is computed as approximately 1.86.
The result exceeds the critical value of 1.833, leading to the rejection of the null hypothesis. The p-value associated with the observed t-value is around 0.048, reinforcing the conclusion that the observed increase is statistically significant at the 0.05 level.
The findings suggest that the campaign had a positive effect on water consumption among adults, with the evidence statistically supporting an increased mean intake. This implies that targeted health campaigns can be effective in modifying hydration behaviors, which is vital for public health initiatives aiming to improve hydration status in populations.
Future research should incorporate larger sample sizes for greater statistical power and consider longitudinal studies to assess sustained behavioral changes over time. Additionally, exploring other factors influencing water consumption, such as climate, activity level, and demographic variables, could offer more comprehensive insights into hydration behaviors and inform more tailored health promotion strategies.
The methodology demonstrated here underscores the importance of rigorous statistical analysis in evaluating public health interventions. Proper hypothesis testing allows researchers and policymakers to confidently interpret behavioral data, ensuring that resource allocation and health messaging are evidence-based and effective for promoting healthier lifestyles.
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