T-Test

The article “Does Smoking Cessation Lead to Weight Gain?” described an experiment in which 322
subjects, selected at random from those show successfully participated in a program to quit smoking
were weighed at the beginning of the program and again one year later . The mean change in weight
was 5.15 lb and the standard deviation was 11.45 lb. Is there sufficient evidence at =.05 to conclude
that the true mean change is positive?

If you are asked to do a t-test on this data, see the following:

t = 5.15/(11.45/√322)

You can probably get away with doing a z-test for this problem because the sample size is so large. However, the formulas are the same for a t or z test.
If you use a z-table for a one-tailed z-test, check the table at .05 level of significance. If the test statistic exceeds the critical or cutoff value from the table, then you conclude there is a difference. If the test statistic does not exceed the critical or cutoff value from the table, then you cannot conclude a difference.

If you use a t-test, determine the degrees of freedom, then check the t-table for a one-tailed test at the .05 level.

I'll let you take it from there.

To determine if there is sufficient evidence to conclude that the true mean change is positive, we can conduct a one-sample t-test.

Step 1: State the hypotheses.
- Null hypothesis (H₀): The true mean change is zero (μ = 0).
- Alternative hypothesis (H₁): The true mean change is positive (μ > 0).

Step 2: Set the significance level (α).
The significance level (α) is given in the question as 0.05.

Step 3: Calculate the test statistic.
The test statistic for a one-sample t-test is calculated as:
t = (sample mean - hypothesized mean) / (standard deviation / square root of sample size)

Given information:
Sample mean (x̄) = 5.15 lb
Standard deviation (s) = 11.45 lb
Sample size (n) = 322
Hypothesized mean (μ₀) = 0 lb

t = (5.15 - 0) / (11.45 / sqrt(322))

Step 4: Find critical value or p-value.
We need to find the critical value or p-value associated with the test statistic to determine the rejection region.

Step 5: Compare the test statistic with the critical value or p-value.
If the test statistic is greater than the critical value or if the p-value is less than the significance level (α), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Step 6: State the conclusion.
Based on the comparison in Step 5, state the conclusion in terms of the original research question.

Unfortunately, the information provided does not include the critical value or p-value associated with the test statistic. To complete the analysis and make a conclusion, we need that information.

To determine whether there is sufficient evidence to conclude that the true mean change is positive, we can conduct a hypothesis test using a T-test.

Here's how you can perform a T-test to answer this question:

Step 1: State the null and alternative hypotheses:
The null hypothesis (H0) states that the true mean change in weight is zero (no change).
The alternative hypothesis (Ha) states that the true mean change in weight is positive.

Step 2: Determine the significance level:
The significance level (alpha) is given as 0.05.

Step 3: Calculate the test statistic:
The T-test statistic, also known as the T-value, can be calculated using the following formula:
T = (sample mean - hypothesized mean) / (standard deviation / √sample size)

In this case, the sample mean change in weight is 5.15 lb, the hypothesized mean change is 0, the standard deviation is 11.45 lb, and the sample size is 322.

T = (5.15 - 0) / (11.45 / √322)
T ≈ 5.15 / (11.45 / √322)
T ≈ 5.15 / (11.45 / 18.00)
T ≈ 5.15 / 0.636
T ≈ 8.10

Step 4: Determine the critical value:
Since the alternative hypothesis is one-tailed (positive), we need to find the critical value for the upper tail. With a sample size of 322 and a significance level of 0.05, we will use a t-distribution table or a statistical software to find the critical value. For a significance level of 0.05 and degrees of freedom (df) equal to 321 (n-1), the critical value is approximately 1.649.

Step 5: Compare the test statistic with the critical value:
If the test statistic (T) is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

In this case, T (8.10) is much greater than the critical value (1.649), indicating that there is strong evidence to reject the null hypothesis.

Step 6: Interpret the result:
Based on the analysis, we can conclude that there is sufficient evidence at a significance level of 0.05 to conclude that the true mean change in weight is positive.