Does the data from the sample support his claim at a significance level of 0.05?
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- Suppose IQ scores were obtained for 20 randomly selected sets of siblings. The 20 pairs of measurements yield x=101.22, y=102.45, r=0.916, P-value=0.000, and y=−8.44+1.1x, where x represents the IQ score of the older child. Find the best predicted value of y given that the older child has an IQ of 99? Use a significance level of 0.05.Use a significance level of a = 9 scores for which x = 3.1 and s = 0.6. Use the traditional method of testing hypotheses. 0.01 to test the claim that u > 2.85. The sample data consist ofSuppose a study reported that the average persin watched 3.37 hours of television per day. a random sample of 15 people gave the number of hours of television watched per day shown below. at the 1% significance level, do the data provide sufficent evidence to conclude that the amount of television watched per day last year by the average person is greater than the value reported in the study? what is the test statistic value for this problem? a.0.28 b.-38 c.none of the above d. 0.58 e.-4.295 f..04 g. 1.94 . the proper conclusion for this problem is? a. reject the null hypothesis we have sufficent evidence to prove the average number of hours people watch tv is more than 3.37 hours. b. Do not reject the null hypothesis we have sufficient evidence to prove the average number of hours of TV watched is greater than 3.37 hours c. Reject the null hypothesis and claim the average number of hours people watch TV is less than 3.37 because the P value is near zero d. Do not reject the null…
- Kenneth, a competitor in cup stacking, claims that his average stacking time is 8.2 seconds. During a practice session, Kenneth has a sample stacking time mean of 7.8 seconds based on 11 trials. At the 4% significance level, does the data provide sufficient evidence to conclude that Kenneth's mean stacking time is less than 8.2 seconds? Accept or reject the hypothesis given the sample data below. H0:μ=8.2 seconds; Ha:μ<8.2 seconds α=0.04 (significance level) z0=−1.75 p=0.0401 Select the correct answer below: a. Do not reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04. b. Reject the null hypothesis because the p-value 0.0401 is greater than the significance level α=0.04. c. Reject the null hypothesis because the value of z is negative. d. Reject the null hypothesis because |−1.75|>0.04. e. Do not reject the null hypothesis because |−1.75|>0.04.You wish to test the following daim (Ha) at a significance level of a = 0.001. H.:P1 2 P2 Ha:P1 < P2 You obtain a sample from the first population with 244 successes and 117 failures. You obtain a sample from the second population with 523 successes and 202 failures. test statistic = [three decimal accuracy] p-value = [four decimal accuracy) The p-value is... O less than (or equal to) a O greater than a This test statistic leads to a decision to... O reject the null hypothesis O fail to reject the null hypothesis As such, the final condusion is that... O There is sufficient evidence to support that the first population proportion is less than the second population proportion. There is not sufficient evidence to support that the first population proportion is less than the second population proportion.The coach of a very popular men’s basketball team claims that the average distance the fans travel to the campus to watch a game is 35 miles. The team members feel otherwise. A sample of 16 fans who travel to games was randomly selected and yielded a mean of M= 36 miles and s= 5 miles. Test the coach’s claim at the 5% (.05) level of significance. one-tailed or two-tailed test: State the hypotheses: df= tα or t value for the critical region = sM = t (test statistic)= Decision:
- The mean potassium content of a popular sports drink is listed as 132 mg in a 32-oz bottle. Analysis of 24 bottles indicates a sample mean of 131.2 mg. (a) State the hypotheses for a two-tailed test of the claimed potassium content.. a. Ho: μ= 132 mg vs. H₁: b. He: ≤132 mg vs. H₁: c. He: ≥132 mg vs. H₁: Oa Ob Oc (b) Assuming a known standard deviation of 2.1 mg, calculate the z test statistic to test the manufacturer's claim. (Round your answer to 2 decimal places. A negative value should be indicated by a minus sign.) Test statistic 132 mg >132 mg <132 mg (c) At the 2 percent level of significance (a = .02) does the sample contradict the manufacturer's claim? Decision Rule: Reject HoAccording to the Statistical Abstract of the United States, the mean family size in 2010 was 3.14 persons. Suppose a random sample of 225 families taken this year yields a sample mean size of 3.05 persons, and suppose we assume the population standard deviation of family sizes is o = 1 person. Using a 5% significance level, test the hypothesis that the mean family size has decreased since 2010. Perform your hypothesis test, and select the best interpretation of your results from the choices below. Select one: O a. At the 5% level of significance the data are statistically significant. We cannot determine the average family size has decreased since 2010. O b. At the 5% level of significance the data are statisticall significant. We can determine the average family size has decreased since 2010. O c. At the 5% level of significance the data is not statistically significant. We cannot conclude the average family size has decreased since 2010. O d. At the 5% level of significance the data…Is the mean length of adult invasive Lionfish in the Atlantic coast of South Carolina greater than 355.6mm? (don't answer this, this is just so you know what the data is for) Answer the following questions and use the provided excel images, data on the far left is lionfish data Question 2. Describe your null and alternative hypotheses using symbols. Question 3. Select the significance level (a). Question 4. Select an appropriate test, calculate the test statistic and p-value using Excel. Question 5. Determine whether you should reject or not reject the null hypothesis based on the p-value. State the reason to support your answer. Question 6. Describe your conclusion in words in the context of your research question. paste data below Use this sheet to perform a 1-sample t-test. 457.2 431.9 COUNT MEAN 20.00 437.45 438.2 STANDARD DEVIATION 16.48 436.4 STANDARD ERROR 3.69 416.8 454.6 Confidence Interval 95% 456.6 Level of Significance 0.05 453.1 404.9 447.9 437.9 HYPOTHESIS TEST t-VALUE…
- Q5. With 0.05 significance level, H1: p > 0.5, and the value of test statistic is z = 1.93. Then theP-value is equal to 0.9732.a. Trueb. FalseA school psychologist was interested in whether students in the chess club scored above or below the mean grade point average (GPA) at her school. She calculated the overall average GPA at the school to be 2.55 with a standard deviation of 0.5 and noted that the distribution was approximately normal. She then figured the average for the 30 students in the chess club. Their mean GPA was 2.76. a. Using the five steps of hypothesis testing, was the group of 30 chess club students' GPA different from students in general at this school? (Use the .05 significance level.) Be sure to sketch the distributions involved. Be sure to fully state the five steps and show ALL calculations/work. b. Additionally, explain the logic of what you did to a person who understands hypothesis testing for studies in which the sample consists of a single individual but is unfamiliar with hypothesis testing involving a sample of more than one individual. (Be sure your explanation includes a discussion of the…