Chapter 9.2 Homework

Which of the following represents an appropriate set of hypotheses?

Ho:μ=0,HA:μ≠0 Explanation A parameter is always tested, the null and alternative hypotheses must be mutually exclusive and collectively exhaustive, and the equal sign must appear in the null hypothesis.

It is advertised that the average braking distance for a small car traveling at 65 miles per hour equals 120 feet. A transportation researcher wants to determine if the statement made in the advertisement is false. She randomly test drives 36 small cars at 65 miles per hour and records the braking distance. The sample average braking distance is computed as 114 feet. Assume that the population standard deviation is 22 feet. (You may find it useful to reference the appropriate table: z table or t table) a. State the null and the alternative hypotheses for the test. b. Calculate the value of the test statistic and the p-value. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Find the critical value(s) if α = 0.01. c. Use α = 0.01 to determine if the average breaking distance differs from 120 feet.

a.) H0: μ = 120; HA: μ ≠ 120 b. ) Given that: n= 36 x̄= 114 σ= 22 μ= 120 z= x̄-μ0/(σ/√n) =114-120/ (22/√36) =-1.64 (Test Value) α/2 = 0.01/2 z= 0.005 (look in the body of the z-table) z= -2.575 C.V.s are z = -2.575 and z = 2.575 > For the two-tail test we have two critical values based on α/2, therefore, the critical values are the cutoff values for the bottom and top 0.005 of the z-values, which are z = -2.575 and z = 2.575 c.) The average breaking distance is not significantly different from 120 feet. >Since the T.S is neither less than -2.575 nor greater than 2.575, T.S. is not in the rejection Region, therefore, we do not reject H0. The average braking distance is not significantly different from 120 feet at 1% significance level.

Access the hourly wage data on the below Excel Data File (Hourly Wage). An economist wants to test if the average hourly wage is less than $22. Assume that the population standard deviation is $6. Click here for the Excel Data File a. Select the null and the alternative hypotheses for the test. b-1. Find the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) b-2. Find the critical value(s) for 5% significant level. c. At α = 0.05, what is the conclusion?

We use the Z-statistic because the standard deviation is KNOWN. a.) H0: μ ≥ 22; HA: μ < 22 b-1.) mean hourly wage= 20.2106 (USING EXCEL) Given that: n= 50 x̄= 20.2106 σ= 6 μ=22 z= x̄-μ/(σ/√n) z= 20.2106-22/ (6/√50) z= -1.79/0.85 z=-2.11 Z Test Statistic: -2.11 b-2.) (we use this equation to find C.V. because population is unknown) 0.05= -1.64 (look in the body of the z-table) C.V. is z = -1.645 >The critical value is the cutoff value of the bottom 5% of the z value, C.V. is z = -1.645. c.) Reject H0; the hourly wage is less than $22. > Since the T.S. is less than -1.645, T.S is in the rejection region, therefore, we reject H0. At the 5% significance level, we conclude that the average hourly wage is less than $22.

Consider the following hypotheses: H0: μ ≤ 12.6HA: μ > 12.6 A sample of 25 observations yields a sample mean of 13.4. Assume that the sample is drawn from a normal population with a population standard deviation of 3.2. (You may find it useful to reference the appropriate table: z table or t table) a-1. Find the Test Statistic and the critical value(s) if α = 0.10. a-2. What is the conclusion if α = 0.10? a-3. Interpret the results at α = 0.10.

a-1. ) Given that: n=25 x̄= 13.4 σ=3.2 μ=12.6 I) z= x̄-μ/(σ/√n) z= 13.4-12.6/ (3.2/√25) z= 0.8/0.64 z= 1.25 Test Statistics= 1.25 (Population is KNOWN) Z= 0.10 (use z-table body to find 0.10) C.V. = -1.28 T.S. = 1.25 and C.V. is z = 1.28 > c.v. is the cut off point for the top 10% of the z-values, z = 1.28 a.2.) Do not reject H0 since the Test Statistic is not greater than the Critical Value EXPLANATION: We do not reject H0 because the T.S. is not grater than the C.V., therefore T.S. is not in the Rejection Region for a right tail test. a-3.) We cannot conclude that the population mean is greater than 12.6.

Customers at Costco spend an average of $130 per trip (The Wall Street Journal, October 6, 2010). One of Costco's rivals would like to determine whether its customers spend more per trip. A survey of the receipts of 25 customers found that the sample mean was $135.25. Assume that the population standard deviation is $10.50 and that spending follows a normal distribution. (You may find it useful to reference the appropriate table: z table or t table) a. Specify the null and alternative hypotheses to test whether average spending at the rival's store is more than $130. b-1.Calculate the value of the test statistic.(Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) b-2.Find the critical value(s) at 5% significance level. c.At the 5% significance level, what is the conclusion to the test? d.Make an inference at α α= 0.05.

a. H0: μ ≤ 130; HA: μ > 130 b-1. The value of the Test Statistic (T.S.) is 2.50 Given that: n= 25 x̄= 135.25 σ= 10.50 μ= 130 z= x̄-μ/(σ/√n) = 135.25-130/ (10.50/√25) = 5.25/2.1 = 2.50 b-2. (Population is KNOWN) 5%= 0.05 (look in the body of the z-table) z= -1.645 C.V. is z = 1.645 >The critical value is the cutoff value of the top 5% of the z values, C.V. is z = 1.645. c. Reject H0 since the T.S. is greater than the C.V. (T.S) 2.50 > (C.V.)-1.645 >Since the T.S. is greater than the C.V. it means the T.S. falls in the rejection region, therefore we reject H0. Customers at this store spend more than $130 at 5% significance level. d. Customers of Costco's rivals spend more than $130 per trip at 5% significance level.

A local bottler in Hawaii wishes to ensure that an average of 16 ounces of passion fruit juice is used to fill each bottle. In order to analyze the accuracy of the bottling process, he takes a random sample of 48 bottles. The mean weight of the passion fruit juice in the sample is 15.80 ounces. Assume that the population standard deviation is 0.8 ounce. (You may find it useful to reference the appropriate table: z table or t table) a. Select the null and the alternative hypotheses to test if the bottling process is inaccurate. b-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) b-2. Find the critical value(s) for 5% significance level. c-1. What is the conclusion at α = 0.05? c-2. Make a recommendation to the bottler.

a.) H0: μ = 16; HA: μ ≠ 16 b-1. ) Given that: n= 48 x̄= 15.80 σ= 0.8 μ= 16 z= x̄-μ/(σ/√n) z= 15.80-16/ (0.8/√48) z= -1.73 Test Statistic: -1.73 b-2.) α/2 =0.05/2 =0.025 (look in the body of the z-table) = -1.96 C.V.s are z = -1.96 and z = 1.96 > The critical values are the cutoff values of the bottom and top 0.025 of the z values, so the C.V.s are z = -1.96 and z = 1.96. c-1.) Do not reject H0 since the T.S. is not in the rejection region. > Since the T.S. is not less than -1.96 nor greater than 1.96, T.S. is not in the rejection region, therefore, we do not reject H0. c-2.) The accuracy of the bottling process is not compromised. > Based on the sample data, the average content of bottles is not significantly different from 16 ounces at the 5% significance level. Thus, the accuracy of the bottling process is not compromised.

According to the Centers for Disease Control and Prevention (February 18, 2016), 1 in 3 American adults don't get enough sleep. A researcher wants to determine if Americans are sleeping less than the recommended 7 hours of sleep on weekdays. He takes a random sample of 150 Americans and computes the average sleep time of 6.7 hours on weekdays. Assume that the population is normally distributed with a known standard deviation of 2.1 hours. Test the researcher's claim at α = 0.01. (You may find it useful to reference the appropriate table: z table or t table) a. Select the relevant null and the alternative hypotheses. b. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) c. Find the critical value(s) at 1% significance level. d. What is the conclusion at α = 0.01? e. Make an inference.

a.) H0: μ ≥ 7; HA: μ < 7 b.) Given that: n= 150 x̄= 6.7 σ= 2.1 μ= 7 z= x̄-μ/(σ/√n) z= 6.7-7/ (2.1/√150) z= -1.75 Test Statistic: -1.75 Convert Z into P-value: (note: we convert the z value by using excel= NORM.S.DIST -1.75,1) Using Excel function =NORM.S.DIST(1.75,1) = 0.0400 c.) 1%=0.01 = -2.33 C.V. is z = -2.33 >The C.V. is the cutoff value of the bottom 1% of the z values, C.V. is z = -2.33. d. p-value(0.0400) < 0.01 Do not reject H0 since the p-value is greater than α. >Since the T.S. is not less than the C.V., the T.S. is not in the rejection region, therefore we do not reject H0. The researcher's claim is not supported by the sample data at 1% significance level. e.) There is insufficient evidence at 1% significance level to suggest that Americans sleep less than the recommended 7 hours of sleep.

Consider the following hypotheses:H0: μ ≥ 150HA: μ < 150A sample of 80 observations results in a sample mean of 144. The population standard deviation is known to be 28. (You may find it useful to reference the appropriate table: z table or t table) a-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) a-2. Find the critical value(s) if α = 0.05 c. Does the above sample evidence enable us to reject the null hypothesis at α = 0.05? d. Interpret the results at α = 0.05.

a.) Test Statistics= -1.92 Given that: n=80 x̄= 144 σ=150 μ=28 z= x̄-μ/(σ/√n) =144-150/ (28/√80) =-6/3.13 =-1.92 b.) α = 0.05 (look in the body of the z-table) =-1.645 C.V. is z = -1.645 > The C.V. is the cutoff value of the bottom 5% of z-values, z = -1.645. c.) Yes since the Test Statistic is less than the critical value. > We reject H0 because the T.S. is less than the critical value, therefore it falls in the Rejection Region for a left tail test.

A university is interested in promoting graduates of its honors program by establishing that the mean GPA of these graduates exceeds 3.50. A sample of 36 honors students is taken and is found to have a mean GPA equal to 3.60. The population standard deviation is assumed to equal 0.40. The parameter to be tested is ___________________________.

the mean GPA of the university honors students