certain drug is used to treat asthma. In a clinical trial of the​ drug, 18 of 266 treated subjects experienced headaches​ (based on data from the​ manufacturer). The accompanying calculator display shows results from a test of the claim that less than 9​% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts​ (a) through​ (e) below.   Is the test​ two-tailed, left-tailed, or​ right-tailed?   What is the test​ statistic?   What is the​ P-value?   What is the null​ hypothesis, and what do you conclude about​ it?   Decide whether to reject the null hypothesis. Choose the correct answer below.

Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 12people who buy insurance from Company A, the mean cost is $153 per month with a standard deviation of $16. For 15 randomly selected customers of Company B, you find that they pay a mean of $160 per month with a standard deviation of $10. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.10 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha: μ1_____μ2 Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places Step 3 of…

Insurance Company A claims that its customers pay less for car insurance, on average, than customers of its competitor, Company B. You wonder if this is true, so you decide to compare the average monthly costs of similar insurance policies from the two companies. For a random sample of 1313 people who buy insurance from Company A, the mean cost is $151$⁢151 per month with a standard deviation of $16$⁢16. For 99 randomly selected customers of Company B, you find that they pay a mean of $158$⁢158 per month with a standard deviation of $19$⁢19. Assume that both populations are approximately normal and that the population variances are equal to test Company A’s claim at the 0.050.05 level of significance. Let customers of Company A be Population 1 and let customers of Company B be Population 2. Step 2 of 3:  Compute the value of the test statistic. Round your answer to three decimal places.

The accompanying table lists pulse rates. Use a 0.05 significance level and apply the methods of two-way analysis of variance. What is the conclusion? Click on the icon to view the data table. C State the null and alternative hypotheses in the test for the effect of an interaction between row and column factors. Ho: There is no interaction between gender and age. H₁: There is an interaction between gender and age. What is the value of the test statistic for this test? F = 3.32 (Round to two decimal places as needed.) What is the corresponding P-value of the test statistic, F, for this P-value= (Round to three decimal places as needed.) Pulse Rates for Gender and Age Over 30 Years of Age Under 30 Years of Age Female 78 103 78 63 61 98 81 98 91 95 77 75 73 65 71 78 61 71 74 55 Female Male 60 81 56 69 68 74 75 68 63 56 46 70 61 65 91 80 59 58 64 59 Male D X

In college, talented runners may join a cross-country team. Runners tend to run their best times when they run even splits. Even splits occur when the runners maintain an even pace throughout the race. The cross-country coach wants to estimate the typical variability in his best runner's 1-mile splits. He takes a random sample of 25 of this runner's mile splits and finds that this runner's mean 1-mile split is 5.44 minutes per mile, with a standard deviation of 0.14 minutes per mile. This runner's 1-mile splits follow a normal distribution. (a) Find the chi-square critical values XL² and Xu² to be used in constructing a 95% confidence interval for the true population standard deviation. (Round your answers to two decimal places.) XL²= XU²= (b) Find the 95% confidence interval for the true variability in his best runner's 1-mile splits. (Round your answers to three decimal places.) lower bound and the upper bound

Perform a left tail test to test whether Machine-2 has less variability. Use a significance level of 0.01. What is the conclusion of the test results?

In a study of birth order and intelligence, IQ tests were given to 18- and 19-year-old men to estimate the size of the difference, if any, between the mean IQs of firstborn sons and secondborn sons. The following data for 10 firstborn sons and 10 secondborn sons are consistent with the means and standard deviations reported in the article. It is reasonable to assume that the samples come from populations that are approximately normal. Can you conclude that the mean IQ of firstborn sons is greater than the mean IQ of secondborn sons? Let μ1 denote the mean IQ of firstborn sons and μ2 denote the mean IQ of secondborn sons. Use the α = 0.01 level and the P-value method with the table. Firstborn 128 101 128 112 121 105 122 98 106 108 Secondborn 121 125 110 107 114 93 80 94 91 83 Part(a): State the appropriate null and alternate hypotheses. H0:   H1: This is a _____…

An intercity bus company is attempting to improve the reliability of its service.  One manager claims that the variability (i.e., standard deviation) in arrivals from the posted schedule at the central bus depot is approximately 3 minutes.  Another manager disagrees, suspecting that the spread in times of arrival is different.  Using variance, what is the population parameter under consideration?  What hypotheses should the managers test in this situation?

A basketball coach believes that the variance of the heights of adult male basketball players is different from the variance of heights for the general population of men. The sample variance of heights, measured in inches, for a random sample of 19 basketball players is 14.22. The sample variance for a random sample of 18 other men is 34.49. Assume that both population distributions are approximately normal and test the coach’s claim using a 0.10 level of significance. Does the evidence support the coach’s claim? Let male basketball players be Population 1 and men in general be Population 2.   Step 1 of 3:  State the null and alternative hypotheses for the test. Fill in the blank below. H0: σ21=σ22: Ha: σ21⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯σ22 Step 2 of 3:  Compute the value of the test statistic. Round your answer to four decimal places. Step 3 of 3:  Draw a conclusion and interpret the decision.

Periodically, customers of a financial services company are asked to evaluate the company's financial consultants and services. Higher ratings on the client satisfaction survey indicate better service, with 7 the maximum service rating. Independent samples of service ratings for two financial consultants are summarized here. Consultant A has 10 years of experience, whereas consultant B has 1 year of experience. Use a = 0.05 and test to see whether the consultant with more experience has the higher population mean service rating. Assume the population variances are unequal. Consultant A Consultant B n1 = 16 n2 10 X1 = 6,81 X2 = 6.25 %3D S1 = 0.68 S2 = 0.75 (a) State the null and alternative hypotheses. O Hoi H1 - H2 = 0 Hai H1- Hz>0 O Ho: H1 - 42 = 0 Hại H1 - H2# 0 O Ho: H1 - H2 > 0 %3D Ha: H1 - H2 = 0 O Ho: H1 - Hz # 0 Hai Hy - Hz = 0 Hoi H1 - H2 = Hai H1- H2 = 0 (b) Compute the value of the test statistic. (Round your answer to three decimal places.) 19.191 (c) What is the p-value?…