Health issues due to being overweight affect all age groups. of children and adolescents 6-11 years of age,18.8% are found to be overweight. A school district randomly sampled 130 in this age group and found that 20 were considered overweight. at Alpha =.05 is this less than the national proportion?
I have worked this out but got a different answer so I need to see if I am doing this correctly.
This is what I have X=263, N=500, Alpha =.05 H0 p=.47 h1 p not = .47this is the claim.
p hat = x/n = 263/500 = .526
z = (.526-.47)/√.526*.474/500)
z = 0.056/√.249324/500)
z = 2.50779
P-value = 0.0121
Think you messed up z=(.526-.47)/sqrt.526*.474/500)it should be Z=(.526-.47)/(.47)(.53)/sqrt 500 = am I right Sir
To determine whether the proportion of overweight children in the given school district is significantly different from the national proportion, we can perform a hypothesis test. Here's how you can approach it:
1. Define the null hypothesis (H0) and alternative hypothesis (H1):
H0: The proportion of overweight children in the school district is equal to the national proportion (p = 0.188).
H1: The proportion of overweight children in the school district is not equal to the national proportion (p ≠ 0.188).
2. Calculate the expected number of overweight children in the school district population based on the national proportion:
Expected number = p * sample size = 0.188 * 130 = 24.44
3. Determine the statistical test to perform:
Since we have a categorical variable (overweight/non-overweight), and we want to compare a single proportion to a known value, we can use a one-sample proportion z-test.
4. Calculate the test statistic (z-score):
z = (Observed proportion - Expected proportion) / Standard error
Observed proportion = Number of overweight children in the sample / Sample size = 20 / 130 = 0.1538
Expected proportion = Proportion based on the national average = 0.188
Standard error = sqrt((p(1-p))/n) = sqrt((0.188*0.812)/130) = 0.0311
z = (0.1538 - 0.188) / 0.0311
5. Find the critical value and p-value:
At α = 0.05 significance level (95% confidence), we need to check if the calculated z-score is within the critical region or if the p-value is less than 0.05.
Find the critical value(s) from the standard normal distribution table or using a statistical software. For a two-tailed test, the critical z-value at α = 0.05 is approximately ±1.96.
Calculate the p-value corresponding to the calculated z-score using the standard normal distribution table or a statistical software.
6. Make a decision:
If the calculated z-score falls within the critical region (|z| > 1.96) or if the p-value is less than the significance level (p < 0.05), we reject the null hypothesis.
Otherwise, if the calculated z-score falls outside the critical region (|z| ≤ 1.96) or if the p-value is greater than the significance level (p ≥ 0.05), we fail to reject the null hypothesis.
Note: Since you mentioned different values (X, N, Alpha), I will assume your calculations are for a different problem. Please update your question with the correct values if you want assistance with your specific calculations.