The screening process for detecting a rare disease is not perfect. Researchers have developed a blood test that is considered fairly reliable. It gives a positive reaction in 98% of the people who have that disease. However, it erroneously gives a positive reaction in 3% of the people who do not have the disease. Answer the following questions using the null hypothesis as "the individual does not have the disease."
1.


a. What is the probability of Type I error? (Round your answer to 2 decimal places.)

Probability


b. What is the probability of Type II error? (Round your answer to 2 decimal places.)

Probability


2.
Consider the following hypotheses:


H0: μ ≤ 12.6
HA: 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 known population standard deviation of 3.2. Use Table 1.



a. Calculate the p-value. (Round "z" value to 2 decimal places and final answer to 4 decimal places.)


p-value



b. What is the conclusion if α = 0.10?


Reject H0 since the p-value is greater than α.

Reject H0 since the p-value is smaller than α.

Do not reject H0 since the p-value is greater than α.

Do not reject H0 since the p-value is smaller than α.



c. Calculate the p-value if the above sample mean was based on a sample of 100 observations. (Round "z" value to 2 decimal places and final answer to 4 decimal places.)


p-value



d. What is the conclusion if α = 0.10?


Reject H0 since the p-value is greater than α.

Reject H0 since the p-value is smaller than α.

Do not reject H0 since the p-value is greater than α.

Do not reject H0 since the p-value is smaller than α.


3.
Consider the following hypotheses:


H0: μ ≥ 150
HA: μ < 150


A sample of 80 observations results in a sample mean of 144. The population standard deviation is known to be 28. Use Table 1.



a. What is the critical value for the test with α = 0.01 and with α = 0.05? (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.)


Critical Value
α = 0.01

α = 0.05

________________________________________


b-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to 4 decimal places. Round your answer to 2 decimal places.)


Test statistic



b-2. Does the above sample evidence enable us to reject the null hypothesis at α = 0.01?


Yes since the value of the test statistic is not less than the negative critical value.

Yes since the value of the test statistic is less than the negative critical value.

No since the value of the test statistic is not less than the negative critical value.

No since the value of the test statistic is less than the negative critical value.



c. Does the above sample evidence enable us to reject the null hypothesis at α = 0.05?


Yes since the value of the test statistic is not less than the negative critical value.

Yes since the value of the test statistic is less than the negative critical value.

No since the value of the test statistic is not less than the negative critical value.

No since the value of the test statistic is less than the negative critical value.

4.

Consider the following hypotheses:


H0: μ = 1,800
HA: μ ≠ 1,800


The population is normally distributed with a population standard deviation of 440. 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. Use Table 1.(Negative values should be indicated by a minus sign. Round intermediate calculations to 4 decimal places. Round "test statistic" values to 2 decimal places and "p-value" to 4 decimal places.)


Test Statistic p-value
a. = 1,850; n = 110



b. = 1,850; n = 280



c. = 1,650; n = 32



d. = 1,700; n = 32



________________________________________rev: 08_21_2013_QC_33738

5.
Consider the following hypotheses:


H0: μ = 120
HA: μ ≠ 120


The population is normally distributed with a population standard deviation of 46. Use Table 1.



a. Use a 5% level of significance to determine the critical value(s) of the test. (Round your answer to 2 decimal places.)


Critical value(s) ±



b-1. Calculate the value of the test statistic with = 132 and n = 50. (Round intermediate calculations to 4 decimal places. Round your answer to 2 decimal places.)


Test statistic



b-2. What is the conclusion at α = 0.05?


Reject H0 since the value of the test statistic is greater than the critical value.

Reject H0 since the value of the test statistic is smaller than the critical value.

Do not reject H0 since the value of the test statistic is greater than the critical value.

Do not reject H0 since the value of the test statistic is smaller than the critical value.



c. Use a 10% level of significance to determine the critical value(s) of the test. (Round your answer to 2 decimal places.)


Critical value(s) ±



d-1. Calculate the value of the test statistic with = 108 and n = 50. (Negative value should be indicated by a minus sign. Round intermediate calculations to 4 decimal places. Round your answer to 2 decimal places.)


Test statistic



d-2. What is the conclusion at α = 0.10?


Reject H0 since the value of the test statistic is not less than the negative critical value.

Reject H0 since the value of the test statistic is less than the negative critical value.

Do not reject H0 since the value of the test statistic is not less than the negative critical value.

Do not reject H0 since the value of the test statistic is less than the negative critical value.

6.
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. Use Table 1.


Use the critical value approach to test the bottler's concern at α = 0.05.


a. Select the null and the alternative hypotheses for the test.


H0: μ ≤ 16; HA: μ > 16

H0: μ ≥ 16; HA: μ < 16

H0: μ = 16; HA: μ ≠ 16



b-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to 4 decimal places. Round your answer to 2 decimal places.)


Test statistic



b-2. Find the critical value(s). (Round your answer to 2 decimal places.)


Critical value(s) ±



b-3. What is the conclusion?


Reject H0 since the value of the test statistic is not less than the negative critical value.

Reject H0 since the value of the test statistic is less than the negative critical value.

Do not reject H0 since the value of the test statistic is not less than the negative critical value.

Do not reject H0 since the value of the test statistic is less than the negative critical value.



c. Make a recommendation to the bottler.


The accuracy of the bottling process is .


7.
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.


Click here for the Excel Data File



a. Select the null and the alternative hypotheses for the test.


H0: μ ≤ 22; HA: μ > 22

H0: μ = 22; HA: μ ≠ 22

H0: μ ≥ 22; HA: μ < 22



b. Use the Excel function Z.TEST to calculate the p-value. Assume that the population standard deviation is $6. (Round your answer to 4 decimal places.)


p-value



c. At α = 0.05 what is the conclusion?


Reject H0; the hourly wage is less than $22.

Do not reject H0; the hourly wage is less than $22.

Do not reject H0; the hourly wage is not less than $22.

Reject H0; the hourly wage is not less than $22.


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  • The screening process for detecting a rare disease is not perfect. Researchers have developed a blood test that is considered fairly reliable. It gives a positive reaction in 98% of the people who …

  • The screening process for detecting a rare disease is not perfect. Researchers have developed a blood test that is considered fairly reliable. It gives a positive reaction in 98% of the people who …