21) Computing the Confidence Intervals for ?d
upper limit = (115.2778 - 114.0556 )+ 3.8325 * ( 1.826046/3) = 3.26
lower limit = (115.2778 - 114.0556 ) - 3.8325 * ( 1.826046/3) = -0.82
Correct option- A) -0.82 < mu <3.26
22) Null hypothesis : The percentage of children who suffer from the disorder is different from 10 %
Alternate hypothesis : The percentage of children who suffer from the disorder is 10 %
Type I error is Probability of rejecting null hypothesis when null hypothesis is true.
Correct option- A) Reject the claim that the percentage of children who suffer from the disorder is different from 10 % when that percentage really is different from 10 %
23) Null hypothesis : the percentage of students that come from single-parents homes is not equal to 11%
Alternate hypothesis : The percentage of students that come from single-parents homes is equal to 11%
Type II error is Probability of accepting null hypothesis when alternate hypothesis is true.
Correct option- D) Reject the claim that the percentage of students that come from single-parents homes is equal to 11% when that percentage is actually 11%
Construct a confidence interval for ?d, the mean of the differences d for the population of...
Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type Il error for the test. 1) The principal of a school claims that the percentage of students at his school that come from single-parent homes is 20%. Identify the type II error for the test. Fail to reject the claim that the percentage of students that come from single- parent homes is equal to 20% when that percentage is actually different from...
Question 17 (4 points) Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type Il error for the test. The principal of a school claims that the percentage of students at his school that come from single-parent homes is 20%. Identify the type Il error for the test. O 1) Fail to reject the claim that the percentage of students that come from single-parent homes is equal to 20% when that percentage...
Construct a confidence interval for pd, the mean of the differences d for the population of paired data. Assume that the population of paired differences is normally distributed. 5) A test of abstract reasoning is given to a random sample of students before and after they 5) completed a formal logic course. The results are given below. Construct a 95% confidence interval for the mean difference between the before and after scores. Before 74 83 75 88 84 63 93...
The principal of a school claims that the percentage of students at his school that come from single-parent homes is 11%. He takes a random sample of 100 students and finds 15 students (15%) come from single-parent homes. At the 0.05 significance level, test his claim by providing each of the following: a. the null and alternative hypothesis b. the test statistic c. the pvalue d. state the final conclusion in nontechnical terms e. describe what type 1 error would...
A.
B.
C.
D.
Construct a confidence interval suitable for testing claim that
students taking non proctored tests get higher mean score than
those taking proctored tests.
___<µ1 - µ2 < ____
Yes/No____ because the confidence interval contains only
positive values/only negative values/zero ______.
E.
Construct a confidence interval suitable for testing claim that
students taking non proctored tests get higher mean score than
those taking proctored tests.
___<µ1 - µ2 < ____
Yes/No____ because the confidence interval contains only...
9.2
question 5
part b only
explain how you find the confidence interval as well
please!!
Homework: 9.2-9.3 homework Save Score: 0.5 of 1 pt 5 of 15 (15 complete) HW Score: 57.81%, 8.67 of 15 pts 9.2.7-T Question Help Proctored Nonproctored A study was done on proctored and nonproctored tests. The results are shown in the table. Assume that the two samples are independent simple 11 H2 random samples selected from normally distributed populations, and do not assume that...
The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below. Type of Household Percent of U.S. Households Observed Number of Households in the Community Married with children 26% 101 Married, no children 29% 118 Single parent 9% 28 One person 25% 97 Other (e.g., roommates, siblings) 11% 67 Use a 5% level of significance to test the claim that the distribution of U.S. households fits the...
The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below. Type of Household Percent of U.S. Households Observed Number of Households in the Community Married with children 26% 100 Married, no children 29% 118 Single parent 9% 30 One person 25% 93 Other (e.g., roommates, siblings) 11% 70 Use a 5% level of significance to test the claim that the distribution of U.S. households fits the...
A bottled water distributor wants to determine whether the mean amount of water contained in 1-gallon bottles purchased from a nationally known water bottling company is actually 1 gallon. You know from the water bottling company specifications that the standard deviation of the amount of water is 0.02 gallon. You select a random sample of 45 bottles, and the mean amount of water per 1-gallon bottle is 0.994 gallon. Complete parts (a) through (d) below. a. Is there evidence that...
The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below. Observed Number of Households in the Community 90 Type of Household Married with children Married, no children Single parent One person Other (e.g., roommates, siblings) Percent of U.S. Households 26% 29% 9% 25% 11% 126 28 100 67 Use a 5% level of significance to test the claim that the distribution of U.S. households fits the...