ECON 5 - Hypothesis Testing

A study by Hewitt Associates showed that 79% of companies offer employees flexible scheduling. Suppose a researcher believes that in accounting firms this figure is lower. The researcher randomly selected 415 accounting firms and through interviews determines that 303 of these firms have flexible scheduling. With a 1% level of significance, does the test show enough evidence to conclude that a significantly lower proportion of accounting firms offer employees flexible scheduling?

-2.99 , reject

A coffee-dispensing machine is supposed to deliver 8 ounces of liquid into each paper cup, but a consumer believes that the actual mean amount is less. The consumer obtained a sample of 16 cups of the dispensed liquid with sampler mean of 7.75 ounces and sample variance of 0.81 ounces. If the dispensed liquid delivered per cup is normally distributed, the appropriate decision at α=0.05 is

Ho: Mu = 8 Ha: Mu!=8 Sample mean = 7.75 Actual mean = 8 Sample variance = .81 n=16 cups alpha=.05 t = (7.75-8)/sqrt(.81/16) = -1.11 which at df = 16-1=15 is not in critical region We therefore, Ans: Fail to reject the 8-ounces claim

The New York Stock Exchange recently reported that the average age of a female shareholder is 44 years. A broker in Chicago wants to know whether this figure is accurate for the female shareholders in Chicago. The broker secures a master list of shareholders in Chicago and takes a random sample of 58 women. Suppose the average age for shareholders in the sample is 45.1 years, with a population standard deviation of 8.7 years. Test to determine whether the broker's sample data differ significantly enough from the 44-years figure released by the New York Stock Exchange to declare that Chicago female shareholders are different in age from female shareholders in general. Use α = 0.05. If no significant difference is noted, what is the broker's probability of committing a Type II error: a. If the average age of a female Chicago shareholder is actually 45 years? b. If the average age of a female Chicago shareholder is actually 46 years? c. If the average age of a female Chicago shareholder is actually 47 years? d. If the average age of a female Chicago shareholder is actually 48 years? e. Construct an OC curve for these data. f. Construct a power curve for these data.

a. .8576 b. .5832 c. 0.2514 d. .0618 e. A f. D

In performing hypothesis tests about the population mean, if the population standard deviation is not known, a t test can be used to test the mean if _______________.

the population is normally distributed

In the testing of hypotheses about a population parameter, in order to increase the power of the test, _________.

the sample size must be increased

A researcher is testing a hypothesis of a single mean. The critical z value for α = .05 and a two-tailed test is ± 1.96. The observed z value from sample data is 2.85. The decision made by the researcher on this information is to ______ the null hypothesis.

Ans: reject Z test statistic = 2.85 > Z critical value = 1.96 Therefore we reject H0 at α = 0.05 The decision made by the researcher based on this information is to reject the null hypothesis.

In the previous problem, if the true alternative mean is 550, then the probability of committing the type II error is: ___________.

0.1977 Using formula 9.1 with the value of Xc computed in the previous problem and u1 which is given in the problem, z1 can be computed. The probability of committing a type II error can be determined with z1 and Table A.5.

A department store wants to use a random sample of 100 customers to test the claim that 40% of its customers browse the store's website prior to visiting the store. If the desired level of significance is 0.05, what are the critical values for the sample proportion that determine the rejection region?

0.304 and 0.496

The power of a hypothesis test is represented by ______________________.

1 - the probability of a Type II error

Ophelia O'Brien, VP of Consumer Credit of American First Banks (AFB), monitors the default rate on personal loans at the AFB member banks. One of her standards is no more than 5% of personal loans should be in default. On each Friday, the default rate is calculated for a sample of 500 personal loans. Last Friday's sample contained 30 defaulted loans. Using α = 0.10, the critical z value is ________.

Ans: 1.28 Since we want to test the claim "no more than 5% of personal loans should be in default." so the null hypothesis is H0: p <= 0.05 and alternative hypothesis are Ha: p > 0.05 Test is right tailed and so critical value of z is 1.28.

A coffee-dispensing machine is supposed to deliver 8 ounces of liquid into each paper cup, but a consumer believes that the actual mean amount is less. The consumer obtained a sample of 49 cups of the dispensed liquid with average of 7.75 ounces. If the sample variance of the dispensed liquid per cup is 0.81 ounces, and α=0.05, the p-value is approximately

Ans: 0.025 Z = (7.75 - 8) / [0.9 / √49]) = -0.25 / (0.9 / 7) = -0.25 / 0.128 = -1.95 Z-score: the test statistic follows the standard normal distribution N(0,1). p-value ≈ 0.025588 Your result is statistically significant: you can reject the null hypothesis and accept the alternative (reject the 8-ounces claim) This decision is made at significance level α = 0.05.

The customer help center in your company receives calls from customers who need help with some of the customized software solutions your company provides. Previous studies had indicated that 20% of customers who call the help center are Hispanics whose native language is Spanish and therefore would prefer to talk to a Spanish-speaking representative. This figure coincides with the national proportion, as shown by multiple larger polls. You want to test the hypothesis that 20% of the callers would prefer to talk to a Spanish-speaking representative. You conduct a statistical study with a sample of 35 calls and find out that 11 of the callers would prefer a Spanish-speaking representative. The significance level for this test is 0.01. The value of the test statistic obtained is:

Ans: 1.69 H0: p = 0.20, Ha: p 0.20 Sample proportion Phat = 11/35 = 0.3143 Test statistics z = Phat - p / sqrt(p(1-p) / n) = 0.3143 - 0.20 / sqrt(0.20 * 0.80 / 35) = 1.69 This is test statistics value. Critical value at 0.01 level is 2.5758. Since test statsistics value is within critical value interval, we do not have sufficient evidence to reject H0. Fail to reject H0. We conclude at 0.01 level that we fail to support the claim.

The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar businesses reported the mean numbers of customers waiting to have their oil changed on Saturday morning is 3.6. Suppose the local oil changing business owner, wants to perform a hypothesis test. The null hypothesis is the population mean is 3.6 and the alternative hypothesis that the population mean is not equal to 3.6. The owner takes a random sample of 16 Saturday mornings during the past year and determines the sample mean is 4.2 and the sample standard deviation is 1.4. It can be assumed that the population is normally distributed. The observed "t" value for this problem is _______.

Ans: 1.71

Eighteen percent of U.S.-based multinational companies provide an allowance for personal long-distance calls for executives living overseas, according to the institute for International Human Resources and the National Foreign Trade Council. Suppose a researcher thinks that U.S.-based multinational companies are having a more difficult time recruiting executives to live overseas and that an increasing number of these companies are providing an allowance for personal long-distance calls to these executives to east the burden of living away from home. To test this hypothesis, a new study is conducted by contacting 376 multinational companies. Twenty-two percent of these surveyed companies are providing an allowance for personal long-distance calls to executives living overseas. Does the test show enough evidence to declare that a significantly higher proportion of multinational companies provide a long-distance call allowance? Let α = .01.

Ans: 2.02 and we fail to reject the null hypothesis Formulating the null and alternative hypotheses, Ho: p <= 0.18 Ha: p > 0.18 As we see, the hypothesized po = 0.18 Getting the point estimate of p, p^, p^ = x / n = 0.22 Getting the standard error of p^, sp, sp = sqrt[po (1 - po)/n] = 0.019812955 Getting the z statistic, z = (p^ - po)/sp = 2.018881066 As this is a 1 tailed test, then, getting the p value,\ p = 0.021749791 significance level = 0.01 As P > 0.01, we FAIL TO REJECT THE NULL HYPOTHESIS. Thus, there is no significant evidence that a significantly higher proportion of multinational companies provide a long-distance call allowance. [CONCLUSION]

The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar businesses reported the mean numbers of customers waiting to have their oil changed on Saturday morning is 3.6. Suppose the local oil changing business owner, wants to perform a hypothesis test. The null hypothesis is the population mean is 3.6 and the alternative hypothesis that the population mean is not equal to 3.6. The owner takes a random sample of 16 Saturday mornings during the past year and determines the sample mean is 4.2 and the sample standard deviation is 1.4. It can be assumed that the population is normally distributed. The table "t" value for this problem is _______.

Ans: 2.131

Discrete Components, Inc. manufactures a line of electrical resistors. Presently, the carbon composition line is producing 100 ohm resistors. The population variance of these resistors must not exceed 4 to conform to industry standards. Periodically, the quality control inspectors check for conformity by randomly select 10 resistors from the line, and calculating the sample variance. The last sample had a variance of 4.36. Assume that the population is normally distributed. Using α = 0.05, the appropriate decision is ___________________.

Ans: fail to reject the null hypothesis Solution: Hypothesis can be written as H0 : (σ^2) <= 4 vs H1 : (σ^2) > 4 The test statistic is chisquare = (n - 1)(s^2)/(σ^2) = (10 - 1) (4.36)/(4) = 9.81 Now , n = 10 So , d.f. = n -1 = 9 α = 0.05 Right tailed test So the critical values is chisquare-α,df Using chi square table , the critical values is 16.92 Test statistic 9.81 is less than 16.92 fail to reject the null hypothesis

Ophelia O'Brien, VP of Consumer Credit of American First Banks (AFB), monitors the default rate on personal loans at the AFB member banks. One of her standards is "no more than 5% of personal loans should be in default." On each Friday, the default rate is calculated for a sample of 500 personal loans. Last Friday's sample contained 30 defaulted loans. Using α = 0.10, the appropriate decision is _______.

Ans: fail to reject the null hypothesis Formulating the null and alternative hypotheses, Ho: p <= 0.05 Ha: p > 0.05 As we see, the hypothesized po = 0.05 Getting the point estimate of p, p^, p^ = x / n = 0.06 Getting the standard error of p^, sp, sp = sqrt[po (1 - po)/n] = 0.009746794 Getting the z statistic, z = (p^ - po)/sp = 1.025978352 As this is a 1 tailed test, then, getting the p value, p = 0.152450894 significance level = 0.1 As P > 0.1, we FAIL TO REJECT THE NULL HYPOTHESIS. Hence, there is no significant evidence that more than 5% of personal loans are in default. [CONCLUSION]

In the testing of hypotheses about the population mean, if the sample size is increased keeping other things like the significance level α same, the critical mean values for the rejected region ___________.

Ans: move closer to the hypothesized value The critical values are determined from the standard error of the sample mean, which is inversely proportional to the sample size. In practical terms, the larger the sample size the more likely the sample mean will fall near the hypothesized population mean (assuming the hypothesis is correct). The threshold for a statistically significant outcome, meaning a sample mean considered as far from the hypothesized value, should therefore be reduced.

Elwin Osbourne, CIO at GFS, Inc., suspects that at least 25% of e-mail messages sent by GFS employees are not business related. A random sample of 300 e-mail messages was selected to test this hypothesis at the 0.01 level of significance. Fifty-four of the messages were not business related. The null hypothesis is ____.

Ans: p = 0.25 Hypotheses: H0: p = 0.25 H1: p < 0.25 Left tailed test So , the null hypothesis is p = 0.25

When using the p-value to test hypotheses, the null hypothesis would be rejected if _________.

Ans: the p-value is less than the significance level α The smaller (closer to 0) the p-value, the stronger is the evidence against the null hypothesis. If the p-value is less than or equal to the specified significance level α, the null hypothesis is rejected; otherwise, the null hypothesis is not rejected. Explanation: In any hypothesis test, we take the decision based on p-value as follows: If the p-value is less than the significance level α, we reject the null hypothesis; and if p-value is greater than the significance level α, we do not reject the null hypothesis. Thus, the null hypothesis is rejected if the p-value is less than the significance level α and the correct option is the fourth option. First option is not correct because p-value can never be negative. Second option is not correct because in any hypothesis test p-value and significance level α are compared to reach a decision. p-value and power of the test are never compared. Third option is not correct because when the p-value is greater than the significance level α, we do not reject the null hypothesis.

In a two-tailed hypothesis about a population mean with a sample size of 100, σ is known, and alpha - 0.10, the rejection region would be _________.

Ans: z < -1.64 and z > 1.64 Alpha represents our rejection region. Since we have two-tailed tests we have two rejection regions. To get rejection on each side (positive and negative side), divide alpha by 2. Hence alpha = 0.10, and a/2 = 0.10 / 2 = 0.05 This means on each side we have 0.05 rejection region. Find the corresponding Z-score. The z-score that will give us an area of 0.05 to its left is between 1.64 and 1.65. Get the average of the z-scores, we have (1.64 +1.65)/2 = 1.645. Since our distribution is symmetrical, its negative counterpart will also give us a rejection region of 0.05. Having two rejection regions of 0.05, we have our alpha 0.10. Means Our Answer Is Correct.

In the testing of hypotheses about the population proportion when the sample size is large enough to satisfy the central limit theorem the critical values are determined using the ________.

Ans: z-distribution (For small values, we use t distribution and for large values we use z-distribution)

An analyst tested the null hypothesis u >= 30 against the alternative hypothesis that u < 30. The analyst reported a p-value of 0.07. For what significance level of alpha will the null hypothesis would be rejected?

The p-value defines the smallest value of alpha for which the null hypothesis can be rejected. Ans: alpha > 0.07

The weight of a USB flash drive is 30 grams and is normally distributed. Periodically, quality control inspectors at Dallas Flash Drives randomly select a sample of 17 USB flash drives. If the mean weight of the USB flash drives is too heavy or too light the machinery is shut down for adjustment; otherwise, the production process continues. The last sample showed a mean and standard deviation of 31.9 and 1.8 grams, respectively. Using α = 0.10, the appropriate decision is _________.

H0: u = 30 Ha: u != 30 Test statistics t = x-u / S / sqrt(n) = 31.9 - 30 / 1.8 / sqrt(17) = 4.352 This is test statistics value. Critical values are -1.746, 1.746 Since test statistics value > 1.746, we have sufficient evidence to reject H0. We conclude at 0.10 level that we have enough evidence to support the claim that mean weight of USB flash drive is too heavy and machinery is shut down for adjustment.

The weight of a USB flash drive is 30 grams and is normally distributed. Periodically, quality control inspectors at Dallas Flash Drives randomly select a sample of 17 USB flash drives. If the mean weight of the USB flash drives is too heavy or too light the machinery is shut down for adjustment; otherwise, the production process continues. The last sample showed a mean and standard deviation of 31.9 and 1.8 grams, respectively. Using α = 0.10, the critical "t" values are _________.

H0: u = 30 Ha: u != 30 Test statistics t = x-u / S / sqrt(n) = 31.9 - 30 / 1.8 / sqrt(17) = 4.352 This is test statistics value. Critical values are -1.746, 1.746 Since test statistics value > 1.746, we have sufficient evidence to reject H0. We conclude at 0.10 level that we have enough evidence to support the claim that mean weight of USB flash drive is too heavy and machinery is shut down for adjustment.

In performing a hypothesis test where the null hypothesis is that the population mean is 4.8 against the alternative hypothesis that the population mean is not equal to 4.8, a random sample of 25 items is selected. The sample mean is 4.1 and the sample standard deviation is 1.4. It can be assumed that the population is normally distributed. The observed "t" value for this problem is _______.

Solution : Given that, u = 4.8 x = 4.1 s = 1.4 n = 25 The null and alternative hypothesis is , H0 : u = 4.8 Ha : u != 4.8 This is the two tailed test . Test statistic = t = (x - u) / s / (square root of n) = (4.1 - 4.8) / 1.4 / (square root of 25) = -2.5 The observed " t " value is -2.5

The rejection region in a hypothesis test depends on ____________.

The rejection region in a hypothesis test depends on the sample size and the sample mean. Ans: the significance level and the type of test (i.e., one-sided or two-sided) The significance level specifies the acceptably low risk of falsely rejecting the null hypothesis if it is actually true. The significance level determines the size of the rejection region. The location of the rejection region depends on the nature of the alternative hypothesis proposed. Test statistics that are very large and/or very small may be considered as supporting the alternative; the rejection region may then appear on the left or right of the hypothesized distribution, or divided between the two sides.

The dean of a business school claims that the average salary of its graduates is more than 85(in $000's). It is known that the population standard deviation is 10 (in $000's). Sample data on the starting salaries of 64 randomly selected recent graduates yielded a mean of 88(in $000's). What is the value of the sample test statistic?

Using formula 9.1 with a u=88, and standard deviation of 10, a sample test statistic can be computed. Ans: 2.40

In a hypothesis test about a population mean with a known population standard deviation, of H0 : u=100 against Ha : u != 100, the sample data yield the test statistic z = 2.17. The p-value for this test is __________.

Using the standard normal table, A.5, we find the probability of obtaining a z value greater than 2.17 is .5000 - .4850 = .0150. Because this is a two-tailed test this value is doubles. Ans: 0.03

The average cost per square foot for office rental space in the central business district of Philadelphia is $23.58, according to Cushman & Wakefield. A large real estate company wants to confirm this figure. The firm conducts a telephone survey of 95 offices in the central business district of Philadelphia and asks the office managers how much they pay in rent per square foot. Suppose the sample average is $22.83 per square foot. The population standard deviation is $5.11. a. Conduct a hypothesis test using α = .05 to determine whether the cost per square foot reported by Cushman & Wakefield should be rejected. b. If the decision in part (a) is to fail to reject and if the actual average cost per square foot is $22.30, what is the probability of committing a Type II error?

a. Fail to reject the null hypothesis b. β = .3156 The Hypotheses is stated below: From the given information, the population mean cost per square foot for office rental space in the central business district of Philadelphia is $23.58, number of surveyed telephone offices are 95, population standard deviation is $5.11, sample mean cost is $22.83 and level of significance is 0.05. Null hypothesis: H0​:μ=23.58 Alternative hypothesis: Ha​:μ=23.58 The null hypothesis states that there is no evidence that cost per square foot reported by Cushman & Wakefield should be rejected. The alternative hypothesis states that there is an evidence that cost per square foot reported by Cushman & Wakefield should be rejected.

Elwin Osbourne, CIO at GFS, Inc., suspects that at least 25% of e-mail messages sent by GFS employees are not business related. A random sample of 300 e-mail messages was selected to test this hypothesis at the 0.01 level of significance. Sixty of the messages were not business related. The appropriate decision is ________.

fail to reject the null hypothesis

The executives of CareFree Insurance, Inc. feel that a majority of our employees perceive a participatory management style at CareFree. A random sample of 200 CareFree employees is selected to test this hypothesis at the 0.05 level of significance. Eighty employees rate the management as participatory. The appropriate decision is ________.

fail to reject the null hypothesis

The executives of CareFree Insurance, Inc. feel that a majority of our employees perceive a participatory management style at CareFree. A random sample of 200 CareFree employees is selected to test this hypothesis at the 0.05 level of significance. Ninety employees rate the management as participatory. The appropriate decision is ________.

fail to reject the null hypothesis

A statistical significant result ___________.

may or may not be substantive

All statistical hypothesis consists of two parts, a null hypothesis and an alternate hypothesis. These two parts must be ___________.

mutually exclusive and collectively exhaustive

Elwin Osbourne, CIO at GFS, Inc., suspects that at least 25% of e-mail messages sent by GFS employees are not business related. A random sample of 300 e-mail messages was selected to test this hypothesis at the 0.01 level of significance. Fifty-four of the messages were not business related. The appropriate decision is ________.

reject the null hypothesis

Ophelia O'Brien, VP of Consumer Credit of American First Banks (AFB), monitors the default rate on personal loans at the AFB member banks. One of her standards is "no more than 5% of personal loans should be in default." On each Friday, the default rate is calculated for a sample of 500 personal loans. Last Friday's sample contained 38 defaulted loans. Using α = 0.10, the appropriate decision is _______.

reject the null hypothesis

Suppose a null hypothesis is that the population mean is greater than or equal to 100. Suppose further that a random sample of 48 items is taken and the population standard deviation is 14. For each of the following α values, compute the probability of committing a Type II error if the population mean actually is 99.

α=.10, β=.7852 α=.05, β=.8749 α=.01, β=.9671 As α gets smaller (other variables remaining constant, β gets larger. Decreasing the probability of committing a Type I error increases the probability of committing a Type II error if other variables are held constant.

Suppose a hypothesis states that the mean is exactly 50. If a random sample of 35 items is taken to test this hypothesis, what is the value of β if the population standard deviation is 7 and the alternative mean is 53? Use α = .01.

β=.5162