Suppose you have computed the following 95% confidence interval for a population mean y from a set of data: (8.56, 11.44). You want to conduct the following hypothesis test using the same data at significance level a = 0.05. HO: H= 12 vs. H 1: µ#12 What conclusion can you draw? O A. As the interval does not contain 12, we reject the null hypothesis. O B. As the interval does not contain 12, we fail to reject the null hypothesis. O C. No conclusion can be drawn since there is no relationship between a confidence interval and a hypothesis test.

Is there a relationship between confidence intervals and two-tailed hypothesis tests? Let c be the level of confidence used to construct a confidence interval from sample data. Let a be the level of significance for a two-tailed hypothesis test. The following statement applies to hypothesis tests of the mean. For a two-tailed hypothesis test with level of significance a and null hypothesis Họ: H = k, we reject Ho whenever k falls outside the c = 1- a confidence interval for u based on the sample data. When k falls within the c = 1 - a confidence interval, we do not reject Ho.- (A corresponding relationship between confidence intervals and two-tailed hypothesis tests also is valid for other parameters, such as p, u, – H2, or p1 - P2, which we will study in later sections.) Whenever the value of k given in the null hypothesis falls outside the c = 1- a confidence interval for the parameter, we reject Hg. For example, consider a two-tailed hypothesis test with a = 0.01 and Ho: H = 21 Hị:…

9.5 Practice Questions Exercise 09.36 (Population Proportion) 1. 2. Consider the following hypothesis test: Ho: p≥ 0.75 Ha: p < 0.75 A sample of 300 items was selected. Compute the p-value and state your conclusion for each of the following sample results. Use a = 0.05. Round your answers to four decimal places. a. p = 0.68 p-value Conclusion: p-value less than or equal to 0.05, reject Ho b. p = 0.72 p-value Conclusion: p-value greater than 0.05, do not reject ✓ Ho c. p = 0.70 p-value Conclusion: p-value less than or equal to 0.05, reject Ho d. p = 0.77 p-value Conclusion: p-value greater than 0.05, do not reject ✓ Ho Icon Key Question 1 01 2 ► Hint(s) Check My Work Hint(s) Check My Work

Tonya wants to estimate what proportion of her school's seniors plan to attend the prom. She interviews an SRS of 50 of the 750 seniors in her school and finds that 36 plan to go to the prom. Which of the following is not one of the conditions that needs to be met in order to construct a confidence interval for the parameter? Large Counts: n(1 – p) = 14 10 Random: The sample is a simple random sample 10% n = 50 is less than 10% of the population size Large Counts: nộ = 36 > 10 Normal/Large Sample: The sample size is large n 50 30. O O O O O

This Question: 4 pts 1 of 9 (0 complete) This Quiz: 45 pts possible Assume a significance level of a = 0.05 and use the given information to complete parts (a) and (b) below. Original claim: More than 54% of adults would erase all of their personal information online if they could. The hypothesis test results in a P-value of 0.1661. a. State a conclusion about the null hypothesis. (Reject H, or fail to reject Ho.) Choose the correct answer below. A. Reject Ho because the P-value is greater than a. B. Reject Ho because the P-value is less than or equal to a. C. Fail to reject H, because the P-value is less than or equal to a. O D. Fail to reject H, because the P-value is greater than a. b. Without using technical terms, state a final conclusion that addresses the original claim. Which of the following is the correct conclusion? A. There is sufficient evidence to support the claim that the percentage of adults that would erase all of their personal information online if they could is…

Given two independent random samples with the following results: n₂ = 8 x₂ = 124 $2 = 28 = 16 n₁ x₁ = 94 $1 = 16 Use this data to find the 90 % confidence interval for the true difference between the population means. Assume that the population variances are equal and that the two populations are normally distributed. Copy Data Step 1 of 3: Find the point estimate that should be used in constructing the confidence interval.

Professor Nord stated that the mean score on the final exam from all the years he has been teaching is a 79%.  Colby was in his most recent class, and his class’s mean score on the final exam was 82%.  Colby decided to run a hypothesis test to determine if the mean score of his class was significantly greater than the mean score of the population.  α = .01. What is the mean score of the population? What is the mean score of the sample? Is this test one-tailed or two-tailed? Why?

One company recently conducted a survey in which 420 women and 420 men were asked "Are you satisfied with the work of the Minister of Health?". Of these, 155 men and 118 women said they were satisfied with the minister's work. Assume that the experimental design was correct. Questions 1 and 2 cover this survey. 1. Find the upper limit of 95% confidence interval for the difference in the proportion of men and women who are happy with the work of the Minister (let men be group 1). 1. 0.03. 2. 0.09. 3. 0.15. 4. It is not possible to apply a normal approach and therefore we can not calculate the confidence interval. 2. What is the rejection area if it is to be examined whether the proportion of men and women who are satisfied with the job of Prime Minister is different (use a = 0.05)? 1. RejectHO efz 1.96. 2. Reject HO if z> 1.64. 3. RejectHO efz> -1.96 or <1.96. 4. Insufficient information provided to answer this.

Based on a random sample of 100 students at Community College in 95% confidence interval for U the mean age of all students at Community College is 21.2 years to 26.6 years ( 21.2<u<26.6). On the basis of this information decide whether you agree or disagree with each of the following statements explain how you decide. A. According to the confidence interval the mean age of all students at Community College cannot possibly be 23.1 years.   B. The 99% confidence interval based on the sample taken above will give a smaller margin of error in the estimate of u.   C. If the sample sizes decrease 240 the 95% confidence interval has a larger margin of error than the one obtained above

Looking at youth depression scores from a general sample of random children in the US, we want to determine if our sample is significantly different than the national average of scores of children with depression in the US. The national average of scores on a measure of depression is 5 points. We have no reason to assume it will be higher or lower than the US national average. The sample:  11 9 12 3 3 4 13 1. What is the null and alternative hypotheses? Do not specify any directionality. 2. What did you discover (using 95% Cl)? Provide an interpretation.