Question 1

For tests of a mean,if other factors are held constant,which statement is correct?

Accepted Answer

In a two-tailed test, the rejection region is split between both tails, making it harder to reject the null hypothesis compared to a one-tailed test where the entire rejection region is in one tail.

Question 2

In testing a proportion,which of the following statements is incorrect?

Accepted Answer

It is not safe to assume normality when the sample proportion is p = .02 and n = 150. The rule of thumb for normality assumption is np(1-p) ≥ 10, which is not satisfied in this case (np(1-p) = 3). Therefore, a different test (such as a chi-squared test) should be used instead of a z-test.

Question 3

A sample of 16 ATM transactions shows a mean transaction time of 67 seconds with a standard deviation of 12 seconds.State the hypotheses to test whether the mean transaction time exceeds 60 seconds.

Accepted Answer

The null hypothesis (H0) is that the mean transaction time does not exceed 60 seconds (μ ≤ 60), and the alternative hypothesis (H1) is that the mean transaction time exceeds 60 seconds (μ > 60). This setup is used to test if the mean time is greater than a specified value.

Question 4

Last year,10 percent of all teenagers purchased a new iPhone.This year,a sample of 260 randomly chosen teenagers showed that 39 had purchased a new iPhone.To test whether the percent has risen,the p-value is approximately:

Accepted Answer

The answer of Last year,10 percent of all teenagers purchased...

Question 5

The process that produces Sonora Bars (a type of candy)is intended to produce bars with a mean weight of 56 gm.The process standard deviation is known to be 0.77 gm.A random sample of 49 candy bars yields a mean weight of 55.82 gm.Find the p-value for a test to see whether the candy bars are smaller than they are supposed to be.

Accepted Answer

To find the p-value for a one-tailed test (since we're testing if the candy bars are smaller than they are supposed to be), we first calculate the z-score using the formula $z = \frac{\bar{x} - \mu}{(\sigma/\sqrt{n})}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values, we get $z = \frac{55.82 - 56}{(0.77/\sqrt{49})} = \frac{-0.18}{0.11} = -1.636$. Looking up the z-score of -1.636 in a standard normal distribution table or using a calculator, we find that the p-value is between .05 and .10, indicating the probability of observing a sample mean of 55.82 gm or less by chance if the true mean is 56 gm.

Question 6

Given that in a one-tail test you cannot reject H₀,can you reject H₀ in a two-tailed test at the same α?

Accepted Answer

In a two-tailed test, the null hypothesis is rejected if the sample statistic falls in either tail of the distribution. Since a one-tailed test only looks at one tail, it is possible that the sample statistic could fall in the other tail and lead to rejection of the null hypothesis in a two-tailed test. However, if the null hypothesis cannot be rejected in a one-tailed test, it cannot be rejected in a two-tailed test at the same significance level because the threshold for rejecting the null hypothesis (the alpha level) is lower in a one-tailed test compared to a two-tailed test. Therefore, the answer is B: No.

Question 7

Last year,10 percent of all teenagers purchased a new iPhone.This year,a sample of 260 randomly chosen teenagers showed that 39 had purchased a new iPhone.The test statistic to find out whether the percent has risen would be:

Accepted Answer

We need to use a one-sample proportion z-test to test whether the proportion of teenagers who purchased a new iPhone has increased from 10% to some other value. The test statistic is calculated as:

z = (p̂ - p) / sqrt(p*(1-p)/n)

where p̂ is the sample proportion, p is the hypothesized population proportion (here, p=0.1), and n is the sample size.

In this case, p̂ = 39/260 = 0.15. Plugging in the values, we get:

z = (0.15 - 0.1) / sqrt(0.1*(1-0.1)/260) = 2.687

The corresponding p-value (assuming a two-tailed test) is 0.007, which is less than the standard cutoff of 0.05. Therefore, we reject the null hypothesis and conclude that there is evidence that the proportion of teenagers who purchased a new iPhone has increased from 10%. The best choice for the test statistic is A, 2.687.

Question 8

For a right-tailed test of a hypothesis for a single population mean with n = 10,the value of the test statistic was t = 1.411.The p-value is:

Accepted Answer

Since this is a right-tailed test, we are interested in finding the area to the right of t. Using a t-table with 9 degrees of freedom (n-1=10-1=9), we find the area to the right of t=1.411 is approximately 0.100. This means the p-value is greater than 0.05 (the significance level commonly used in hypothesis testing). Since the p-value is between 0.10 and 0.05, the best choice is B.

Question 9

When testing the hypothesis H₀: μ = 100 with n = 100 and σ² = 100,we find that the sample mean is 97.The test statistic is:

Accepted Answer

To find the test statistic, we use the formula:

t = (sample mean - hypothesized mean) / (standard deviation of sample mean)

t = (97 - 100) / (100 / sqrt(100))

t = -3.00

Therefore, the answer is A) -3.000.

Question 10

Last year,10 percent of all teenagers purchased a new iPhone.This year,a sample of 260 randomly chosen teenagers showed that 39 had purchased a new iPhone.To test whether the percent has risen,the critical value at α = .05 is:

Accepted Answer

The test is a one-tailed z-test for proportions. At α = 0.05, the critical value for a one-tailed test is 1.645.

Question 11

The process that produces Sonora Bars (a type of candy)is intended to produce bars with a mean weight of 56 gm.The process standard deviation is known to be 0.77 gm.A random sample of 49 candy bars yields a mean weight of 55.82 gm.Which are the hypotheses to test whether the mean is smaller than it is supposed to be?

Accepted Answer

The null hypothesis (H0) is that the mean weight of Sonora Bars is equal to 56 gm. The alternative hypothesis (Ha) is that the mean weight of Sonora Bars is less than 56 gm. Since we want to test if the mean is smaller than it is supposed to be, we need to use a one-tailed test with the alternative hypothesis being μ < 56. Therefore, the correct choice is B, which has the hypothesis H₀: μ ≥ 56 (null) and H₁: μ < 56 (alternative).

Question 12

Which of the following is not a characteristic of the t distribution?

Accepted Answer

The t distribution is similar to the z distribution when n is large, not when n is small. When n is small, the t distribution is wider and flatter than the z distribution, with more probability in the tails.

Question 13

Given H₀: μ ≥ 18 and H₁: μ < 18,we would commit Type I error if we:

Accepted Answer

Type I error occurs when we reject a true null hypothesis. In this case, the null hypothesis is that μ ≥ 18. If we conclude that μ < 18 based on the sample data, we are rejecting the null hypothesis, which is only true if μ < 18. Therefore, if the true value of μ is actually greater than or equal to 18, we have committed a Type I error by rejecting the null hypothesis.

Question 14

For a sample size of n = 100,and σ = 10,we want to test the hypothesis H₀: μ = 100.The sample mean is 103.The test statistic is:

Accepted Answer

To find the test statistic, we use the formula:
z = (x - μ) / (σ / sqrt(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.

Plugging in the values, we get:
z = (103 - 100) / (10 / sqrt(100)) = 3

Since the alternative hypothesis is not given, we assume a two-tailed test with a significance level of α = 0.05.

Looking at the z-table, the z-value that corresponds to a cumulative area of 0.025 in each tail is approximately 1.96.

Since our calculated z-value of 3 is greater than 1.96, we reject the null hypothesis and conclude that there is significant evidence to suggest that the true population mean is not equal to 100.

Question 15

A sample of 16 ATM transactions shows a mean transaction time of 67 seconds with a standard deviation of 12 seconds.Find the critical value to test whether the mean transaction time exceeds 60 seconds at α = .01.

Accepted Answer

The answer of A sample of 16 ATM transactions shows...

Question 16

A sample of 16 ATM transactions shows a mean transaction time of 67 seconds with a standard deviation of 12 seconds.Find the test statistic to decide whether the mean transaction time exceeds 60 seconds.

Accepted Answer

The test statistic is calculated using the formula $t = \frac{\bar{x} - \mu}{(s/\sqrt{n})}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $s$ is the sample standard deviation, and $n$ is the sample size. Plugging in the given values: $t = \frac{67 - 60}{(12/\sqrt{16})} = \frac{7}{3} = 2.333$.

Question 17

The process that produces Sonora Bars (a type of candy)is intended to produce bars with a mean weight of 56 gm.The process standard deviation is known to be 0.77 gm.A random sample of 49 candy bars yields a mean weight of 55.82 gm.Find the test statistic to see whether the candy bars are smaller than they are supposed to be.

Accepted Answer

We need to conduct a one-sample t-test to see if the sample mean of 55.82 gm is significantly different from the population mean of 56 gm. 
The formula for the t-test is:
t = (x̅ - μ) / (s / √n)
Where x̅ is the sample mean, μ is the population mean, s is the population standard deviation, and n is the sample size. 
Plugging in the given values, we get:
t = (55.82 - 56) / (0.77 / √49) = -1.636
Comparing this to a t-distribution table with 48 degrees of freedom (49-1), at a significance level of 0.05, the critical value is 1.677. Since our test statistic (-1.636) falls within the rejection region (less than -1.677), we can reject the null hypothesis and conclude that the candy bars are smaller than they are supposed to be.

Question 18

Given a normal distribution with σ = 3,we want to test the hypothesis H₀: μ = 20.We find that the sample mean is 21.The test statistic is:

Accepted Answer

The answer of Given a normal distribution with σ =...

Question 19

Which of the following is not a valid null hypothesis?

Accepted Answer

A null hypothesis specifies there is no effect or no difference, and it is usually formulated as an equality (e.g., μ = 0). The option stating μ ≠ 0 is an alternative hypothesis, not a null hypothesis, because it specifies a difference or effect.

Question 20

Ajax Peanut Butter's quality control allows 2 percent of the jars to exceed the quality standard for insect fragments.A sample of 150 jars from the current day's production reveals that 30 exceed the quality standard for insect fragments.Which is incorrect?

Accepted Answer

The answer of Ajax Peanut Butter's quality control allows 2...

Quiz 9: One-Sample Hypothesis Tests