CHAPTER 9.1 Homework

What is the decision rule when using the p-value approach to hypothesis testing?

Reject Ho if the p-value < α.

Under the assumption that the null hypothesis is true as an equality, the p-value is the likelihood of observing a sample mean that is at least as extreme as the one derived from the given sample.

TRUE

When conducting a hypothesis test concerning the population mean, and the population standard deviation is known, the value of the test statistic is calculated as _________.

A

Rates on 30-year fixed mortgages continue to be at historic lows (Chron Business News, September 23, 2010). According to Freddie Mac, the average rate for 30-year fixed loans for the week was 4.37%. An economist wants to test if there is any change in the mortgage rates in the following week. She searches the Internet for 30-year fixed loans in the following week and reports the rates offered by seven banks as 4.25%, 4.125%, 4.375%, 4.50%, 4.75%, 4.375%, and 4.875%. Assume that rates are normally distributed. (You may find it useful to reference the appropriate table: z table or t table) a. Select the appropriate hypotheses to test if the average mortgage rate differs from 4.37%. b-1. Calculate the value of the test statistic? (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.) b-2. Find the p-value. c. At the 5% significance level, does the average mortgage rate differ from 4.37%?

Given: 1)Calculating the mean(x̄ ) x̄ = 4.25+4.125+ 4.375+4.50+ 4.75+ 4.375+4.875/7 = 31.25/7 x̄ =4.4643 2) Calculating the standard deviation (σ) σ = √(xi - x̄)^2/ (N - 1) = (4.25)^2 + (4.125)^2 + (4.375)^2 + (4.50)^2 + (4.75)^2 + (4.375)^2 + (4.875)^2 / 7-1 = 0.4286/6 =0.0714 σ= √0.0714 = 0.2673 [https://www.calculator.net/standard-deviation-calculator.html?numberinputs=4.25%2C+4.125%2C+4.375%2C+4.50%2C+4.75%2C+4.375%2C+4.875&ctype=s&x=97&y=32] -------------------------------------------------- a. H0: μ = 4.37; HA: μ ≠ 4.37 b-1. TS= ( x̄ - µ ) / ( σ/√(n) TS= (4.4643-4.37)/ (0.2673/√7) Test Statistic= 0.933 b-2. p-value ≥ 0.10 c. Do not reject H0; the claim is not supported by the data.

H0: μ = 7,300 HA: μ ≠ 7,300 The population is normally distributed with a population standard deviation of 700. Compute the value of the test statistic and the resulting p-value for each of the following sample results. For each sample, determine if you can "reject/do not reject" the null hypothesis at the 10% significance level. (You may find it useful to reference the appropriate table: z table or t table) (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round "test statistic" values to 2 decimal places and "p-value" to 4 decimal places.)

Part a) Test Statistic :-Z = ( X - µ ) / ( σ / √(n))Z = ( 7340 - 7300 ) / ( 700 / √( 125 ))Z = 0.64 P value = 2 * P ( Z > 0.64 ) = 2 * 1 - P ( Z < 0.64 ) = 0.5229 Looking for the value Z = 0.64 in standard normal table to find the P value Reject null hypothesis if P value < α = 0.1 level of significanceSince 0.5229 > 0.1 ,hence we reject null hypothesis Result :- We fail to reject null hypothesis Part b) Test Statistic :-Z = ( x̄ - µ ) / ( σ / √(n))Z = ( 7340 - 7300 ) / ( 700 / √( 205 ))Z = 0.82 P value = 2 * P ( Z > 0.82 ) = 2 * 1 - P ( Z < 0.82 ) = 0.4133 Looking for the value Z = 0.82 in standard normal table to find the P value Decision based on P valueReject null hypothesis if P value < α = 0.1 level of significanceSince 0.4133 > 0.1 ,hence we reject null hypothesis Result : We fail to reject null hypothesis Part c) Test Statistic :-Z = ( x̄ - µ ) / ( σ / √(n))Z = ( 7020 - 7300 ) / ( 700 / √( 33 ))Z = -2.30 P value = 2 * P ( Z > 2.30 ) = 2 * 1 - P ( Z < 2.30 ) = 0.0216Looking for the value Z = -2.30 in standard normal table to find the P value Decision based on P valueReject null hypothesis if P value < α = 0.1 level of significanceSince 0.0216 < 0.1 ,hence we reject null hypothesis Result :- Reject null hypothesis Part d) Test Statistic :-Z = ( x̄ - µ ) / ( σ / √(n))Z = ( 7110 - 7300 ) / ( 700 / √( 33 ))Z = -1.56 P value = 2 * P ( Z > 1.56 ) = 2 * 1 - P ( Z < 1.56 ) = 0.1189 Looking for the value Z = -1.56 in standard normal table to find the P value Decision based on P valueReject null hypothesis if P value < α = 0.1 level of significanceSince 0.1189 > 0.1 ,hence we reject null hypothesis Result :- We fail to reject null hypothesis

A hypothesis test regarding the population mean µ is based on the sampling distribution of the sample mean X̄ (x bar)

TRUE

Consider the following hypotheses: H0: μ ≤ 36.9HA: μ > 36.9 A sample of 49 observations yields a sample mean of 38.1. Assume that the sample is drawn from a normal population with a population standard deviation of 4.2. (You may find it useful to reference the appropriate table: z table or t table) a-1. Find the p-value. a-2. What is the conclusion if α = 0.10? a-3. Interpret the results at α = 0.10. b-1. Calculate the p-value if the above sample mean was based on a sample of 130 observations. b-2. What is the conclusion if α = 0.10? b-3. Interpret the results at α = 0.10.

a-1) 0.01 p-value < 0.025 Finding the P-value: Z = (x̄- µ ) / ( σ / √(n)) Z= [(38.1-36.9)/(4.2/√49)] P(Z)> 2= 0.97725 P(Z)=1-0.97725 P(Z)= 0.02275 a-2) Reject H0 since the p-value is smaller than α. a-3) We conclude that the population mean is greater than 36.9. b-1) p-value < 0.01 b-2) Reject H0 since the p-value is smaller than α. b-3) We conclude that the population mean is greater than 36.9

A realtor in Mission Viejo, California, believes that the average price of a house is more than $500 thousand. Click here for the Excel Data File a. Select the null and the alternative hypotheses for the test. b-1. Assume the population standard deviation is $100 (in $1,000s). What is 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 p-value. c. At α = 0.05, what is the conclusion?

a. H0: μ ≤ 500,000; HA: μ > 500,000 b-1. Test Statistics: 0.96 b-2. p value≥0.10 c. Do not reject H0; realtor's claim is not supported by the data. {check picture for b-1 solution}

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 $25. Assume that the population standard deviation is $8. 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 p-value. c. At α = 0.05, what is the conclusion?

a. H0: μ ≥ 25; HA: μ < 25 b-1. Test Statistic: -1.64 GIVEN: the sum of all hourly wage (x̄ )= 1157.18 the total number of samples (n)= 50 1157.18/50= 23.1436 µ= 25 σ= 8 z= x̄ - µ/ (σ/n) 23.1436/25/(8/√50)= -1.64 ; where Z =0.0505 b-2. 0.05≤ p-value < 0.10 c. Do not reject H0, the hourly wage is not less than 25

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

If the null hypothesis is rejected at a 1% significance level, then _____________.

the null hypothesis will be rejected at a 5% significance level

A hypothesis test regarding the population mean is based on ________________________________.

the sampling distribution of the sample mean

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 value of the test statistic is ____________.

z = 1.50 u=3.5 n=36 x=3.6 s=0.4 test statistic, Z = ( x̄ - µ ) / ( σ / √(n)) z = 1.5