. The state of Maryland has a healthy children goal for childhood obesity. In the next 5 years, the state has a goal of decreasing the obesity rate to 11.3%. The current population mean obesity rate is 15.6%. After two years of a pilot intervention program in Garrett County Maryland that involves 25 children, the county has measured a mean obesity rate of 14.2%. The standard deviation for this rate is 1.1%. What is the value of the test statistic and is this a significant change at a 0.01 level?

a) Statistic = -6.37, no it is not significant
b) Statistic = -6.37, yes it is significant
c) Statistic = -19.55, no it is not significant
d) Statistic = -19.55, yes it is significant
e) Cannot determine based on information provided

I used a t-test and got A) -6.37, not significant

Using a t-test I agree with -6.37. However, it is significant because -6.37 is a large t-value.

I would never use a t-test when dealing with % which are equivalent to proportions. I would have done a 1 proportion z-test... but given the choices that you were given it looks like your teacher was using a t-test.

To determine the test statistic, we can use a dependent samples t-test. This test is appropriate in this scenario because we are comparing the mean obesity rate before and after the intervention program for the same group of children in Garrett County.

Here are the steps to calculate the test statistic:

1. Define the null and alternative hypotheses:
- Null hypothesis (H0): There is no significant change in the mean obesity rate before and after the intervention program.
- Alternative hypothesis (Ha): There is a significant change in the mean obesity rate before and after the intervention program.

2. Determine the significance level: In this case, the significance level is given as 0.01.

3. Calculate the test statistic using the formula:
T = (X̄d - μd) / (s / √n),
where X̄d is the mean difference, μd is the hypothesized mean difference, s is the standard deviation of the sample, and n is the sample size.

In this case, X̄d = 14.2% - 15.6% = -1.4%, μd = 0 (since there is no significant change in the null hypothesis), s = 1.1%, and n = 25.

Plugging in the values, we get:
T = (-1.4% - 0) / (1.1% / √25)
= -1.4% / (1.1% / 5)
= -1.4% / 0.22%

Converting the percentages to decimals, we have:
T = (-0.014 / 0.011) * 100
= -1.27

4. Determine the critical value: Since the sample size is small (n = 25), we need to look up the critical value for a t-distribution with n-1 degrees of freedom and a significance level of 0.01. Looking up in a t-table or using software, the critical value is approximately -2.797.

5. Compare the test statistic with the critical value: If the test statistic is less than the critical value (in absolute value), we reject the null hypothesis.

In this case, |-1.27| < |-2.797|, so we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that there is a significant change in the mean obesity rate before and after the intervention program.

Therefore, the correct answer is a) Statistic = -6.37, no, it is not significant.