Episodes of dehydration are common at outdoors music festivals. At a particular large annual event, a first
aid team has recorded over the years 85 cases on average per day. In 2018, during a weekend they observe 78, 86,
79, and 81 cases. Assuming the distributions of cases is normal, can you confirm that the number of cases has
significantly dropped in 2018? Assume alpha=0.10
Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to your Z score.
I'll let you do the calculations.
To confirm if the number of cases has significantly dropped in 2018, we can perform a hypothesis test using the given data. Let's define our null and alternative hypotheses:
Null Hypothesis (H0): The number of cases in 2018 is equal to the average of 85 cases per day.
Alternative Hypothesis (H1): The number of cases in 2018 has significantly dropped.
We can use a one-sample t-test to perform this hypothesis test. Here are the steps to follow:
Step 1: Calculate the sample mean and sample standard deviation of the numbers of cases in 2018.
Sample mean (x̄) = (78 + 86 + 79 + 81) / 4 = 82
Sample standard deviation (s) = √[((78 - 82)² + (86 - 82)² + (79 - 82)² + (81 - 82)²) / 3] = √[10.67] ≈ 3.27
Step 2: Define the significance level (alpha) as 0.10.
Step 3: Calculate the test statistic (t-statistic) using the formula:
t = (x̄ - μ) / (s / √n)
where μ is the hypothesized mean (85), s is the sample standard deviation, and n is the sample size (4).
t = (82 - 85) / (3.27 / √4) ≈ -0.92
Step 4: Determine the critical value for the given alpha and degrees of freedom.
Since we're assuming the distribution is normal, we will use a t-distribution. With 3 degrees of freedom and alpha = 0.10, the critical value is approximately ±1.895.
Step 5: Compare the calculated t-statistic to the critical value.
|-0.92| < 1.895, so we fail to reject the null hypothesis.
Step 6: Interpret the result.
The analysis suggests that there is not enough evidence to conclude that the number of cases significantly dropped in 2018. The observed data is not significantly different from the average of 85 cases per day, based on a 10% significance level.
In conclusion, we cannot confirm that the number of cases has significantly dropped in 2018 based on the given data and hypothesis test with an alpha of 0.10.