The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) μ and standard deviation σ = 0.1. The brand advertises that the mean nicotine content of their cigarettes is 1.5, but measurements on a random sample of 400 cigarettes of this brand gave a mean of x =1.52. Is this evidence that the mean nicotine

content is actually higher than advertised? To answer this, test the hypotheses of
H0: μ = 1.5 vs. Ha: μ > 1.5
at a significance level of α = 0.01.

1. The test statistic for this test is
A) z = -4.00
B) z = -0.20
C) z = 0.20
D) z = 4.00

2. Based on the p-value of the test and the given significance level, what would you
conclude?
A) Fail to reject H0, indicating evidence that the mean nicotine content in this brand of
cigarettes equals 1.5 milligrams.
B) Reject H0, indicating evidence that the mean nicotine content in this brand of
cigarettes is greater than 1.5 milligrams.
C) There is a 5% chance that the null hypothesis is true.
D) We cannot make a conclusion here since we do not know the true mean of the
population.

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 related to the Z score to answer 2.

To test the hypotheses, we need to calculate the test statistic and the p-value.

1. The test statistic for this test is:

To calculate the test statistic, we use the formula:

z = (x - μ) / (σ / √n)

where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

Given:
x = 1.52 (sample mean)
μ = 1.5 (population mean)
σ = 0.1 (population standard deviation)
n = 400 (sample size)

Plugging in the values, we get:

z = (1.52 - 1.5) / (0.1 / √400)
= (0.02) / (0.1 / 20)
= 0.02 / 0.005
= 4

Therefore, the test statistic for this test is z = 4.00. Thus, the correct answer is D.

2. Based on the p-value of the test and the given significance level, we would conclude:

To find the p-value, we compare the test statistic to the critical value. Since the alternative hypothesis is μ > 1.5, it is a one-tailed test.

At a significance level of α = 0.01, the critical value is z = 2.33.

Since 4.00 > 2.33, the p-value is less than 0.01. This means that the probability of obtaining a sample mean of 1.52 or higher, assuming the null hypothesis is true (μ = 1.5), is less than 0.01.

Therefore, based on the p-value being less than the significance level, we reject the null hypothesis (H0: μ = 1.5) and conclude that there is evidence to suggest that the mean nicotine content is actually higher than advertised. Hence, the correct answer is B.

To answer the first question, we need to calculate the test statistic for this hypothesis test. The formula for the test statistic is:

z = (x - μ) / (σ / √n)

where x is the sample mean, μ is the hypothesized population mean, σ is the population standard deviation, and n is the sample size.

From the given information, x = 1.52, μ = 1.5, σ = 0.1, and n = 400.

Plugging these values into the formula, we get:

z = (1.52 - 1.5) / (0.1 / √400)
= 0.02 / (0.1 / 20)
= 0.02 / 0.005
= 4

Thus, the test statistic for this test is z = 4.00. So the answer to the first question is D) z = 4.00.

To answer the second question, we need to compare the p-value of the test to the given significance level of α = 0.01. The p-value is the probability of observing a test statistic as extreme as the one calculated (or more extreme) under the assumption that the null hypothesis is true. In this case, since our alternative hypothesis is μ > 1.5, we are interested in the right-tailed test.

Using the test statistic of z = 4.00, we can calculate the p-value using a standard normal distribution table or a statistical software. The p-value for a z-statistic of 4.00 in a right-tailed test is extremely small, much smaller than 0.01.

Since the p-value is smaller than the significance level, we can reject the null hypothesis. Therefore, the answer to the second question is B) Reject H0, indicating evidence that the mean nicotine content in this brand of cigarettes is greater than 1.5 milligrams.

As for the third question, none of the provided answer choices address the likelihood of the null hypothesis being true. The conclusion from the hypothesis test is that there is strong evidence against the null hypothesis, not a specific probability assigned to it. Therefore, the correct answer is not among the choices provided.