In one study of smokers who tried to quit smoking with nicotine patch therapy, 39 were smoking one year after the treatment, and 32 were not smoking one year after the treatment. Use a 0.10 significance level to test the claim that among the smokers who try to quit with the nicotine patch therapy, the majority are smoking a year after the treatment. Do these results suggest that the nicotine patch therapy is ineffective?

Take a shot, what do you think?
Hint: your question involves a binominal distribution; the formula for the standard deviation is sqrt(n*p*q), where n is the sample size, p it the expected probability, and q=1-p.
Also, you will need a cumulative normal distribution table, probably in the back of your stats text.

To test the claim, we will perform a hypothesis test.

Step 1: State the null hypothesis (H₀) and alternative hypothesis (H₁).
H₀: p ≤ 0.5 (The proportion of smokers who continue smoking one year after treatment is less than or equal to 50%)
H₁: p > 0.5 (The proportion of smokers who continue smoking one year after treatment is greater than 50%)

Step 2: Find the test statistic.
We will use the formula for the test statistic in this case:
z = (p̂ - p) / √(p(1 - p) / n)
where p̂ is the sample proportion, p is the hypothesized proportion, and n is the sample size.

In this case, p̂ = 39/71, p = 0.5, and n = 71. Plug these values into the formula:

z = (39/71 - 0.5) / √(0.5(1 - 0.5) / 71)
z ≈ 0.548 / 0.059
z ≈ 9.29

Step 3: Find the p-value.
Using a standard normal table or calculator, we can find the corresponding p-value for our test statistic, z = 9.29.

p-value ≈ 0

Step 4: Make a decision.
Since our p-value is less than our significance level (α = 0.10), we reject the null hypothesis.

Conclusion:
These results suggest that among the smokers who try to quit with the nicotine patch therapy, the majority are smoking a year after the treatment. However, this does not necessarily mean that the nicotine patch therapy is completely ineffective, as there might be other factors affecting the result. Further research is needed to determine the effectiveness of the therapy.

To test the claim, we need to perform a hypothesis test. The null hypothesis (H0) is that the majority of smokers who try to quit with nicotine patch therapy are smoking a year after treatment. The alternative hypothesis (Ha) is that the majority are not smoking a year after treatment.

Let's define p as the true proportion of smokers who are smoking a year after treatment. We will use p-hat as the sample proportion.

Given that 39 smokers are smoking a year after treatment, out of a total of 39 + 32 = 71 smokers, the sample proportion is p-hat = 39/71 ≈ 0.5493.

To test the claim, we will use a one-sample proportion z-test. The test statistic (z-score) is calculated using the formula:

z = (p-hat - p) / sqrt( p * (1 - p) / n )

Where n is the sample size.

In this case, we want to test if the majority are smoking. To do this, we need to check if p is less than 0.5.

Let's proceed to calculate the test statistic.

To test the claim that the majority of smokers are smoking a year after the treatment using a 0.10 significance level, we can use a one-sample proportion hypothesis test.

First, let's define our null and alternative hypotheses:
- The null hypothesis (H0) is that the proportion of smokers who are smoking a year after the treatment is equal to or less than 50% (no majority).
- The alternative hypothesis (Ha) is that the proportion of smokers who are smoking a year after the treatment is greater than 50% (majority).

Now, let's calculate the test statistic. We will use the sample proportion as an estimate for the true population proportion.

Sample proportion (p̂) = number of smokers not smoking a year after / total sample size
= 32 / (32 + 39)
≈ 0.450

Next, we calculate the standard deviation (σ) using the formula sqrt(n * p * q). Here, n is the total sample size, p is the hypothesized proportion, and q = 1 - p.

Standard deviation (σ) = sqrt((32 + 39) * 0.5 * (1 - 0.5))
= sqrt(71 * 0.5 * 0.5)
≈ 4.23

Now, we can calculate the test statistic (z-score) using the formula:

z = (p̂ - p) / σ

Where p is the hypothesized proportion (0.5) and σ is the standard deviation.

z = (0.450 - 0.5) / 4.23
≈ -0.118

To determine the critical value for a one-tailed test (since we are testing whether the proportion is greater than 0.5), we need to look up the z-critical value from the cumulative normal distribution table. At a 0.10 significance level, the critical value will be approximately 1.28.

Since the obtained test statistic (-0.118) is not greater than the critical value (1.28), we fail to reject the null hypothesis. Therefore, we do not have enough evidence to suggest that the nicotine patch therapy is ineffective.