MATH 110 Module 7 Exam (Latest, 2024-2025) / MATH110 Module 7 Exam : Portage Learning, Exams of Nursing

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MATH110 Module 7 Exam/ MATH 110 Statistics

Module 7 Exam/ MATH110 Statistics Module 7 Exam:

Portage Learning

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 3

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.

Ho: u = 85 H1: u < 85 ---> left tailed because LESS THAN u = 85 n = 35 xbar = 82

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.

P(Z< p=""><>

This corresponds to z= - 1..

H1: u < 1020 a = 0.04 ---- > look on standard normal distribution chart for left, 0.04006 is closest value to 0.04, z =

Left tailed test z = - 1.

Answer Key

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

Suppose p represents the probability that a person is unemployed H 0 : p=.04. H 1 : p>.04. Since this is a right-tailed test, we must find the z that satisfies P(Z>z)=.02. In the standard normal table, we find that z.. 02 =2.05. This is a right-tailed test, we reject the null hypothesis if the z-score is greater than 2.05.

Recall that for proportions, the mean and standard deviation are found by:

Since this is a right-tailed test, and the z-score is greater than 2.05, we reject the null hypothesis.