Question: Question: The following data are paired by date. Let x and y be random variables representing wind direction at 5 a.m. and 5 p.m., respectively

Question:

The following data are paired by date. Let x and y be random variables representing wind direction at 5 a.m. and 5 p.m., respectively (units are degrees on a compass, with 0 representing true north). The readings were taken at seeding level in a cloud seeding experiment. A random sample of days gave the following information.x1711401972244617525772172y1401422171255724521835147

x21826511019318019094890y2032181001702451171409960Use the sign test with a 5% level of significance to test the claim that the distributions of wind directions at 5 a.m. and 5 p.m. are different. Interpret the results.

What is the level of significance?

State the null and alternate hypotheses.

H0: Distributions are the same. H1: Distributions are different.

H0: Distributions are different. H1: Distributions are different.

H0: Distributions are the same. H1: Distributions are the same.

H0: Distributions are different. H1: Distributions are the same.

Compute the sample test statistic. (Round your answer to two decimal places.)

What sampling distribution will you use?

normal

Student's t

chi-square

uniform

Find the P-value of the sample test statistic. (Round your answer to four decimal places.)

Interpret your conclusion in the context of the application.

Fail to reject the null hypothesis, there is insufficient evidence that the wind directions at 5 a.m. and 5 p.m. are different.

Fail to reject the null hypothesis, there is sufficient evidence that the wind directions at 5 a.m. and 5 p.m. are different.

Reject the null hypothesis, there is insufficient evidence that the wind directions at 5 a.m. and 5 p.m. are different.

Reject the null hypothesis, there is sufficient evidence that the wind directions at 5 a.m. and 5 p.m. are different.

Question:The following data are paired by date. Let x and y berandom variables representing wind direction at 5 a.m. and 5 p.m., respectively(units are degrees on a compass, with 0 representing true north). Thereadings were taken at seeding level in a cloud seeding experiment. A

Use a Kolmogorov Smirnov (KS) Test in order to determine if the following data set comes from an Exponential distribution with mean equal to five Data 0.433577647 1.077296386 1.461024528 2.037106422 3.671167985 3.724253017 3.815970293 3.905489821 6.842680422 6.9339538392. The duration of calls in minutes over a telephone line is as follows: 14.03 6.42 25.56 17.56 3.25 7.80 3.83 1.31 11.50 6.43 6.98 8.97 9.68 10.03 5.72 Use Kolmogorov-Smirnov test with a = 0.05 to verify that the numbers given have an Exponential Distribution with 1=0.11. a- State the hypothesis. b- Test the data by using K-S test.Repeated measures ANOVA and one-way ANOVA; what's the difference between them? Select the best answer. repeated measures ANOVA is for between subjects designs that contain more than three groups. repeated measures ANOVA is for within subjects designs that look at changes in a variable over time. one way ANOVAs are for within subjects designs normally and repeated measures ANOVA is only for longitudinal designs repeated measures ANOVA is the same thing as a one way ANOVAA general social survey questioned people on their happiness and familyr income. The data are shown in Table 1. Table 1: Frequencies in happiness and income categories Income Not too happy Pretty happy Very happy Total Above average H 159' 1111] 290 Average 53 3?? 2.11 E46 Below average 94 249 83 426 TUI'AL 1&8 F30 414 1362 {a} Is this an example of a randomised experiment or an observaijonal study? Explain your answer. [bi 1liilhat proportion of people with Above average income are 1very happy? [c] What proportion of people who are Not too happy have Below average income? {d} 1Ilv'hat proportion of people have Average income and are Not too happy? Thank you

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