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Questions and answers related to hypothesis testing, including the definition of null and alternative hypotheses, Type I and Type II errors, left-tailed, right-tailed, and two-tailed tests, and the calculation of z-scores for hypothesis testing. The questions are related to real-world scenarios, such as the circulation of a newspaper, the vitamin C intake of pregnant women, and the unemployment rate in a city. useful for students studying statistics and hypothesis testing in mathematics and related fields.
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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.
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.
H1: u < 1020 a = 0.04- -> look on standard normal distribution chart for left, 0.04006 is closest value to 0.04, z = -1. Left tailed test z = -1.
Suppose that we have a problem for which the null and alternative hypothesis are given by: H 0 : μ=1020. H 1 :μ<
Is this a right-tailed test, left-tailed test, or two-tailed test. Find the z value based on a
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
o = 16 a = 0.025 ---> look on standard normal distribution chart left side = -1.96 z = -1. oxbar = ( o / √n ) ( 16 / √35 ) = 2.70449 ≈ 2. z score = ( (xbar - u ) / oxbar ) ( ( 82 - 85) / 2.70) = -1.11111 ≈ -1.
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.
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.
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%