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MATH 110 Module 7 Exam
a) Define the null and alternative hypothesis for the following. Also, explain what it would mean to make a Type I error and explain what it would mean to make a Type II error.
Exam Page 1
a) Define the null and alternative hypothesis for the following. Also, explain what it would mean to make a Type I error and explain what it would mean to make a Type II error.
The newspaper in a certain city had a circulation of 19,000 per day in 2010. You believe that the newspaper’s circulation is different than 19,000 today.
Ho: u = 19000 H1: u ≠ 19000
Type I Error: Reject the null hypothesis saying the mean newspaper circulation in a certain city is 19000 per day when the mean IS 19000 per day. Type II Error: Accept the null hypothesis saying the mean newspaper circulation in a certain city is 19000 per day when the mean it IS DIFFERENT than 19000 per day.
b) Define the null and alternative hypothesis for the following. Also, explain what it would mean to make a Type I error and explain what it would mean to make a Type II error.
A certain website had 6000 hits per month a year ago. You believe that the number of hits per month is less than that today.
Ho: u = 6000 H1: u < 6000
Type I Error: Reject the null hypothesis saying a certain website had mean number of 6000 hits per month a year ago when the mean number of hit DID have 6000 hits per month. Type II Error: Accept the null hypothesis saying a certain website had a mean number of 6000 hits per month a year ago when the mean number of hit was LESS THAN 6000 hits per month.
Answer Key
Exam Page 2
Suppose that we have a problem for which the null and alternative hypothesis are given by:
H 0 : μ=1020. H 1 :μ< 1020.
Is this a right-tailed test, left-tailed test, or two-tailed test. Find the z value based on a level of significance of .04.
Left tailed test because the alternative hypothesis predicts it will be less than 1020.
Ho: u = 1020
The newspaper in a certain city had a circulation of 19,000 per day in 2010. You believe that the newspaper’s circulation is different than 19,000 today.
a) H 0 : μ=19,000. H 1 :μ≠19,000.
Type I error: Reject the null hypothesis that the circulation today is 19,000 when the circulation today actually is 19,000.
Type II error: Do not reject the null hypothesis when the circulation is different than 19,000.
b) Define the null and alternative hypothesis for the following. Also, explain what it would mean to make a Type I error and explain what it would mean to make a Type II error.
A certain website had 6000 hits per month a year ago. You believe that the number of hits per month is less than that today.
b) H 0 : μ=6000. H 1 :μ<6000.
Type I error: Reject the null hypothesis that the website gets 6000 hits per month when the website gets at least 6000 hits per month.
Type II error: Do not reject the null hypothesis when the number of hits per month is less than 6000.
It is recommended that pregnant women over eighteen years old get 85 milligrams of vitamin C each day. A doctor is concerned that her pregnant patients are not getting enough vitamin C. So, she collects data on 35 of her patients and finds that the mean vitamin intake of these 35 patients is 82 milligrams per day with a standard deviation of 16 milligrams per day. Based on a level of significance of α = .025, test the hypothesis.
H 0 : μ=85 milligrams per day. H 1 : μ< 85 milligrams per day.
This is a left-tailed test, so we must find a z that satisfies P(Z<="" p="">
We find the z-score:
Notice that since the z-score is greater than - 1.96, we do not reject the null hypothesis.
o = 16 a = 0.025 ---> look on standard normal distribution chart left side = - 1. z = - 1.
oxbar = ( o / √n ) ( 16 / √35 ) = 2.70449 ≈ 2.
z score = ( (xbar - u ) / oxbar ) ( ( 82 - 85) / 2.70) = - 1.11111 ≈ - 1.
Z score = - 1.11. Z score is greater than - 1.96. We accept the null hypothesis and reject the alternative hypothesis. There is sufficient amount of evidence to support the claim that pregnant patients are getting at least 85 milligrams of vitamin C each day.
Answer Key
A mayor claims that the unemployment rate in his city is 4 %. Many people think that the unemployment rate is higher. So, 95 residents of the city are contacted and it is found that 8 of them are unemployed. Can the mayor’s claim be supported to a level of significance of α = .02, test the hypothesis.
Exam Page 4
A mayor claims that the unemployment rate in his city is 4 %. Many people think that the unemployment rate is higher. So, 95 residents of the city are contacted and it is found that 8 of them are unemployed. Can the mayor’s claim be supported to a level of significance of α = .02, test the hypothesis.
Ho: p = 0. H1: p > 0.04 ---> right tailed because greater than p = 0. n = 95 xbar = 8 a = 0.02 ---> subtract from 1 because H1 is greater than (1 - 0.02) = 0.98 ---- > look on standard normal distribution chart right side, 0.97982 closest value to
z = 2.
u = (np) (95 x 0.04) = 3.
os = √ (np)(1-p) √(3.8)(1-0.04) = 1.90997 ≈ 1.
z score = ( (xbar - u) / os) ( ( 8 - 3.8) / 1.91 ) = 2.19895 ≈ 2.
Z score = 2.20. 2.20 is greater than 2.05. We reject the null hypothesis and accept the alternative hypothesis. There is sufficient amount of evidence to support the claim that the unemployment rate in the mayor's city is higher than 4%
Answer Key
