Ch. 10 Homework

The average cost of tuition and room and board at a small private liberal arts college is reported to be $8,500 per term, but a financial administrator believes that the average cost is higher. A study conducted using 350 small liberal arts colleges showed that the average cost per term is $8,745. The population standard deviation is $1,200. Let α = 0.05. What is the critical z value for this test? +1.960 -1.960 +1.645 -1.645

+1.645. The null and alternate hypotheses are H0: µ ≤ $8,500 and H1: µ > $8,500. This is a one (right)-tailed test; the population standard deviation is known, and α = 0.05, so the critical value of z is +1.645.

What are the critical z-values for a two-tailed hypothesis test if α = 0.01? ±1.960 ±2.326 ±2.576 ±1.645

±2.576 For a two-tailed hypothesis test with a 0.01 significance level, the critical z values are ±2.576.

A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. The null hypothesis is ______. H0: µ = 6.6 H0: µ ≥ 6.6 H0: µ > 7.6 H0: µ ≤ 7.6

H0: µ = 6.6 The null hypothesis is that the population mean is 6.6 pounds, written H0: µ = 6.6.

A null hypothesis makes a claim about a __________. Population parameter Sample statistic Sample mean Type II error

Population parameter A null hypothesis is a statement about the value of a population parameter.

For a null hypothesis, H0: µ = 4,000, if the 1% level of significance is used and the z-test statistic is +6.00, what is our decision regarding the null hypothesis? Do not reject H0. Reject H0. Reject H1. None Apply.

Reject H0. For a two-tailed test and α = 0.01, the z critical values are -2.576 and +2.576. Since the z-test statistic is +6.00, it is greater than +2.576, and the decision is to reject the null hypothesis.

A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. What is the sample standard deviation? 1.177 1.188 1.386 1.090

1.177 The sample standard deviation is 1.177, found by √∑(x-x̅)²/n-1

If the alternate hypothesis states that µ ≠ 4,000, where is the rejection region for the hypothesis test? In both tails In the lower or left tail In the upper or right tail In the center

In both tails. If the alternate hypothesis is an inequality, we use a two-tailed test. The rejection region is in both the upper and lower tails of the distribution.

To conduct a test of hypothesis with a small sample, we make an assumption that __________. A larger computed value of t will be needed to reject the null hypothesis The region of acceptance will be wider than for large samples The confidence interval will be wider than for large samples The population is normally distributed

The population is normally distributed. The t statistic is used to test a hypothesis about a population mean. The t test requires that the population be normally distributed.

The average cost of tuition and room and board at a small private liberal arts college is reported to be $8,500 per term, but a financial administrator believes that the average cost is higher. A study conducted using 350 small liberal arts colleges showed that the average cost per term is $8,745. The population standard deviation is $1,200. Let α = 0.05. What is the test statistic for this test? ±3.82 +0.204 -3.82 +3.82

+3.82. The null and alternate hypotheses are H0: µ ≤ $8,500 and H1: µ > $8,500. This is a one (right)-tailed test; the population standard deviation is known. Based on the sample information, the Z= (x̄-μ)/(σ/√n)= (8,745-8,500)/(1,200/√350)= +3.82

The average cost of tuition and room and board at small private liberal arts colleges is reported to be $8,500 per term, but a financial administrator believes that the average cost is higher. A study conducted using 350 small liberal arts colleges showed that the average cost per term is $8,745. The population standard deviation is $1,200. Let α = 0.05. What is the p-value for this test? 0.0000 0.0124 0.0500 0.4938

0.0000. H0: µ ≤ $8,500 and H1: µ > $8,500. The p-value is the probability of observing a sample value as extreme as, or more extreme than, the value observed, given that the null hypothesis is true. The probability of getting a sample mean of $8,745 or greater, assuming a population mean of $8,500, corresponds to the probability of obtaining a z value greater than 3.82. This probability is beyond the range of the "areas under the normal curve" table, so the probability is extremely small or virtually zero.

A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. What is the sample variance? 1.177 6.6 1.385 7.6

1.385 The sample variance is computed as ∑(x-x̅)²/n-1 and is 1.386, computed by summing the squared deviations from the mean (8.317) and then dividing by the number of degrees of freedom (6).

If the critical z value for a hypothesis test equals 2.45, what value of the test statistic would provide the least chance of making a Type I error? 3.74 10,000 2.46 4.56

10,000 Recall a Type I error occurs if the null hypothesis is rejected when it should not be rejected. An extremely large value of the test statistic, such as z = 10,000, indicates that the difference between the hypothesized population mean and the sample mean is very large. Therefore, given the sample evidence, it is highly likely that the null hypothesis is false and should be rejected with a very small probability of a Type I error.

A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. What are the degrees of freedom? 7 8 6 6.6

6 The problem does not state the population standard deviation, only sample information. Therefore, to test the null hypothesis, we must use the t statistic with its associated degrees of freedom, n - 1. There are 6 degrees of freedom, found by 7 - 1.

A machine is set to fill the small-size packages of M&M candies with 56 candies per bag. A sample revealed: three bags of 56, two bags of 57, one bag of 55, and two bags of 58. To test the hypothesis that the mean candies per bag is 56, how many degrees of freedom are there? 9 1 8 7

7 8 bags, or 3 + 2 + 1 + 2, were sampled. Therefore, n = 8, and the degrees of freedom n - 1 = 7.

A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. What is the sample mean? 6.6 7.6 1.177 2.447

7.6 The sample mean is 7.557, rounded to 7.6. It is found by summing all the values (52.9) and dividing by the sample size (7).

What are the critical values for a two-tailed test with a 0.01 level of significance when n is large and the population standard deviation is known? Above 1.960 and below -1.960 Above 1.645 and below -1.645 Above 2.576 and below -2.576 Above 1.000 and below -1.000

Above 2.576 and below -2.576. In hypothesis testing for the mean when the population standard deviation is known, the z test statistic is employed. For a two-tailed test and α = 0.01, the critical values are -2.576 and 2.576

In hypothesis testing, what is the level of significance? The risk of rejecting the null hypothesis when it is true. A value symbolized by the Greek letter α. A value between 0 and 1. It is selected before a decision rule can be formulated. All apply.

All apply.

A random sample of size 15 is selected from a normal population. The population standard deviation is unknown. Assume the null hypothesis indicates a two-tailed test and the researcher decided to use the 0.10 significance level. For what values of t will the null hypothesis not be rejected? To the left of -1.282 or to the right of 1.282 To the left of -1.345 or to the right of 1.345 Between -1.761 and 1.761 To the left of -1.645 or to the right of 1.645

Between -1.761 and 1.761 This is a two-tailed test, α = 0.10, and the population standard deviation is unknown. The sample size is 15, so the degrees of freedom are 15 - 1 = 14, and the t critical values are -1.761 and +1.761. We fail to reject the null hypothesis if the computed t is between -1.761 and +1.761.

The mean annual income of certified welders is normally distributed with a mean of $50,000 and a population standard deviation of $2,000. The ship building association wishes to find out whether their welders earn more or less than $50,000 annually. The alternate hypothesis is that the mean is not $50,000. If the level of significance is 0.10, what is the decision rule? Do not reject the null hypothesis if computed z lies between -1.645 and +1.645; otherwise, reject it. Do not reject the null hypothesis if computed z is greater than 1.645; otherwise, reject it. Do not reject the null hypothesis if computed z lies between -1.960 and +1.960; otherwise, reject it. Reject the null hypothesis if computed z is below -1.960; otherwise, reject it.

Do not reject the null hypothesis if computed z lies between -1.645 and +1.645; otherwise, reject it. H0: μ = 50,000; H1: μ ≠ 50,000. Since the population standard deviation is known, the z critical values for a two-tailed test at the 0.10 significance level are ± 1.645. So the decision rule is: Do not reject the null if the computed z statistic is between -1.645 and 1.645.

A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. What is the decision for a statistical significant change in average weights at birth at the 5% level of significance? Fail to reject the null hypothesis. Reject the null hypothesis and conclude the mean is higher than 6.6 lb. Reject the null hypothesis and conclude the mean is lower than 6.6 lb. Cannot calculate because the population standard deviation is unknown.

Fail to reject the null hypothesis. H0: µ = 6.6; H1: µ ≠ 6.6. The sample mean is 7.6 and the sample standard deviation is 1.177. The t critical values for a two-tailed test at the 0.05 significance level with 6 degrees of freedom are ±2.447. The t test statistic is t= (x̄-μ)/(s/√n)= (7.6-6.60/(1.177/√7)= 2.08.The computed t statistic is not in the rejection regions, so we fail to reject the null hypothesis.

What is a Type II error? Failing to reject a false null hypothesis Rejecting a false null hypothesis Accepting a false alternate hypothesis Rejecting a false alternate hypothesis

Failing to reject a false null hypothesis. A Type II error is not rejecting the null hypothesis when it is false.

The average cost of tuition and room and board at small private liberal arts colleges is reported to be $8,500 per term, but a financial administrator believes that the average cost is higher. A study conducted using 350 small liberal arts colleges showed that the average cost per term is $8,745. The population standard deviation is $1,200. Let α = 0.05. Based on the computed test statistic or p-value, what is our decision about the average cost? Equal to $8,500 Greater than $8,500 Less than $8,500 Not equal to $8,500

Greater than $8,500. The null and alternate hypotheses are H0: µ ≤ $8,500 and H1: µ > $8,500. This is a one (right)-tailed test; the population standard deviation is known. Based on the sample information, the Z= (x̄-μ)/(σ/√n)= (8,745-8,500)/(1,200/√350)= + 3.82. The p-value is the probability of observing a sample value as extreme as, or more extreme than, the value observed, given that the null hypothesis is true. The probability of getting a sample mean of $8,745 or greater, assuming a population mean of $8,500 corresponds to the probability of obtaining a z value greater than 3.82. This probability is beyond the range of the "areas under the normal curve" table, so the probability is extremely small or virtually zero. The p-value, 0.0000, is less than the significance level 0.05, so the decision is to reject the null hypothesis and conclude the mean is greater than $8,500.

For a two-tailed test with a 0.05 significance level, where is the rejection region when n is large and the population standard deviation is known? Between ±1.960 Between ±1.645 Greater than +1.960 and less than -1.960 Greater than +1.645 and less than -1.645

Greater than +1.960 and less than -1.960. In hypothesis testing for the mean when the population standard deviation is known, the z test statistic is employed. For a two-tailed test and α = 0.05, the critical values are -1.960 and +1.960. The rejection region is z values less than -1.960 and z values greater than +1.960.

For an alternative hypothesis: µ > 6,700, where is the rejection region for the hypothesis test located? In both tails In the left or lower tail In the right or upper tail In the center

In the right or upper tail. Since the alternative hypothesis is that the mean is greater than a specific value, we have a one-tailed test with the entire rejection region in the right or upper tail. We will only reject H0: µ ≤ 6,700 if a sample mean is significantly greater than 6,700 and those values will be in the right or upper tail.

Define the level of significance. It is the probability of a Type II error. It is the probability of a Type I error. It is a z value of 1.96. It is the beta error.

It is the probability of a Type I error. The level of significance is the probability of a Type I error, or the likelihood that a null hypothesis will be rejected when it is true

A consumer products company wants to increase the absorption capacity of a sponge. Based on past data, the average sponge could absorb 3.5 ounces. After the redesign, the absorption amounts of a sample of sponges were (in ounces): 4.1, 3.7, 3.3, 3.5, 3.8, 3.9, 3.6, 3.8, 4.0, and 3.9. At the 0.01 level of significance, what is the decision rule to test if the new design increased the absorption of the sponge? Reject null hypothesis if computed t is less than 2.580. Reject null hypothesis if computed t is greater than 2.821. Reject null hypothesis if computed z is 1.96 or larger. Reject null hypothesis if computed t is less than 2.764.

Reject null hypothesis if computed t is greater than 2.821. This is a one (right)-tailed test, α = 0.01, and the population standard deviation is unknown. The sample size is 10, so the degrees of freedom are 10 - 1 = 9 and the t critical value is +2.821. Reject the null hypothesis if the computed t is greater than 2.821.

The mean weight of newborn infants at a community hospital is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. Does the sample data show a significant increase in the average birthrate at a 5% level of significance? Fail to reject the null hypothesis and conclude the mean is 6.6 lb. Reject the null hypothesis and conclude the mean is lower than 6.6 lb. Reject the null hypothesis and conclude the mean is greater than 6.6 lb. Cannot calculate because the population standard deviation is unknown.

Reject the null hypothesis and conclude the mean is greater than 6.6 lb. H0: µ = 6.6; H1: µ ≠ 6.6. Note that the problem states a one (right)-tailed alternate hypothesis, H1: μ > 6.6. The t critical value for a one-tailed test at the 0.05 significance level with 6 degrees of freedom is +1.943. The t test statistic is t= (x̄-μ)/(s/√n)= (7.6-6.6)/(1.177/√7)= +2.081. The computed t statistic is in the rejection region, so we reject the null hypothesis and conclude that there has been an increase in the mean weight.

What is another name for the alternate hypothesis? Null hypothesis Hypothesis of no difference Rejected hypothesis Research hypothesis

Research hypothesis. The alternate hypothesis describes what you will conclude if you reject the null hypothesis. It is also referred to as the research hypothesis. The alternate hypothesis is accepted if the sample data provide us with enough statistical evidence to reject the null hypothesis.

A manufacturer wants to increase the shelf life of a line of cake mixes. Past records indicate that the average shelf life of the mix is 216 days. After a revised mix has been developed, a sample of nine boxes of cake mix gave these shelf lives (in days): 215, 217, 218, 219, 216, 217, 217, 218, and 218. Using α = 0.025, has the shelf life of the cake mix increased? Yes, because computed t is greater than the critical value. Yes, because computed t is less than the critical value. No, because computed t lies in the region of acceptance. No, because 217.24 is quite close to 216.

Yes, because computed t is greater than the critical value. This is a one (right)-tailed test, α = 0.025, and the population standard deviation is unknown. The sample size is 9, so the degrees of freedom are 9 - 1 = 8 and the critical value is +2.306. Based on the sample information, t= (x̄-μ)/(s/√n)= (217.22-216)/(1.20/√9)= 3.05. The t statistic is greater than 2.306. Therefore, the null hypothesis is rejected and the shelf life has increased.

The mean length of a candy bar is 43 millimeters. There is concern that the settings of the machine cutting the bars have changed. Test the claim at the 0.02 level that there has been no change in the mean length. The alternate hypothesis is that there has been a change. Twelve bars (n = 12) were selected at random and their lengths recorded. The lengths are (in millimeters) 42, 39, 42, 45, 43, 40, 39, 41, 40, 42, 43, and 42. The mean of the sample is 41.5 and the standard deviation is 1.784. Computed t = -2.913. Has there been a statistically significant change in the mean length of the bars? Yes, because the computed t lies in the rejection region. No, because the information given is not complete. No, because the computed t lies in the area to the right of -2.718. Yes, because 43 is greater than 41.5.

Yes, because the computed t lies in the rejection region. This is a two-tailed test, α = 0.02, and the population standard deviation is unknown. The sample size is 12, so the degrees of freedom are 12 - 1 = 11 and the critical values of t are -2.718 and 2.718. Based on the sample information, t= (x̄-μ)/(s/√n)= (41.5-43)/(1.784/√12)= -2.913. The t statistic is less than -2.718, therefore the null hypothesis is rejected and there has been a change in the length of the bars.

The mean annual incomes of certified welders are normally distributed with the mean of $50,000 and a standard deviation of $2,000. The ship building association wishes to find out whether their welders earn more or less than $50,000 annually. The alternate hypothesis is that the mean is not $50,000. If the level of significance is 0.10, what is the critical value? +1.645 -1.282 ±1.282 ±1.645

±1.645 H0: μ = $50,000; H1: μ ≠ $50,000. This a two-tailed hypothesis test of a population mean with α = 0.10. The critical z values are -1.645 and +1.645.

A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. If α = 0.05, what is the critical t value? -2.365 ±1.96 ±2.365 ±2.447

±2.447 H0: µ = 6.6; H1: µ ≠ 6.6. The t critical values for a two-tailed test at the 0.05 significance level with n - 1, or 6 degrees of freedom are ±2.447.

For a one-tailed hypothesis test, the critical z value of the test statistic is -2.33. Which of the following is true about the hypothesis test? α = 0.05 for a lower-tailed test α = 0.01 for a lower-tailed test α = 0.05 for an upper-tailed test α = 0.01 for an upper-tailed test

α = 0.01 for a lower-tailed test. For a one-tailed test and consulting the "areas under the normal curve" for z = -2.33, the area in the tail is 0.5000 - 0.4901 = 0.0099, that rounds to a probability of 0.01. Since the critical value is negative, the rejection region is in the lower tail.