buddy is working with a 2x5 connection board and intends to analyze it with a chi- square test. How many degrees of freedom does the distribution of the test size have if the null hypothesis is true? 1. 2 2. 4 3. 5 4. 10
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Q: 0.9554
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A: As per our guidelines we are suppose to do first question only
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- Need help urgently pleaseTest the claim that the proportion of people who own a pet is smaller than 50% at the 0.05 significance level. The null and alternative hypothesis would be: Но: р 3 0.5 Но:д > 0.5 Но:д 0.5 Ні:р 0.5 Hi:р + 0.5 The test is: left-tailed two-tailed right-tailed Based on a sample of 600 people, 288 owned a pet, the p-value is: |(to 4 decimals) Based on this we: Fail to reject the null hypothesis Reject the null hypothesisGiven that a-0.01%, find the Zeit if the hypothesis testing is a two tailed test. O a. Zcrit = 3.8906 and -3.8906 O b. Zerit 1.2816 and -1.2816 O c. Zcrit = 3.0902 and -3.0902 O d. Zcrit = 1.6449 and -1.6449
- hopes that the results of a trial period will enable her to conclude with a level of significance of 0.05 that the new procedure increases the average profit of installing a carpet. What will be the consequences if she makes a Type I error on her hypothesis test? A Type I error would mean that Carpetland implements a new procedure that actually decreases profits. A Type I error would mean that Carpetland implements a new procedure that actually increases profits. A Type 1 error would mean that Carpetland does not implement a new procedure that increases profits. A Type I error would mean that Carpetland does not implement a new procedure that decreases profits.. A Type I error would mean that Carpetland implements a new procedure that actually does not increase profits. A Type I error would mean that Carpetland implements a new procedure that actually does not decrease profits.Does college major depend on gender? Researchers ask a group of college students about their majors. They also group the students by their gender. The data are recorded in the contingency table below, and a chi-square Test of Independence at the 5% significance level is performed. Arts Humanities Sciences Row Total Men 11 5 19 35 Women 11 22 9 42 Column Total 22 27 28 77 (a) The null and alternative hypotheses are: H0: The two variables are independent, so gender does not affect college major. Ha: The two variables are dependent, so gender does affect college major. (b) Compute the test statistic, rounded to one decimal place. (Use expected frequencies that are also rounded to one decimal place.)(c) Calculate the test statistic. d) Decide whether to reject or fail to reject the null hypothesis. Then interpret the decision in the context of the original claim.
- Test the claim that the proportion of people who own cats is smaller than 30% at the 0.025 significance level. The null and alternative hypothesis would be: Ho: p=0.3 Ho: p = 0.3 Ho: p= 0.3 H₁: p 0.3 The p-value is: Ho: p = 0.3 Ho: p = 0.3 H₁: p0.3 H₁:p> 0.3 (to 2 decimals) (to 2 decimals)Test the claim that the mean GPA of night students is larger than 2.3 at the .025 significance level. The null and alternative hypothesis would be: Но : р — 2.3 Но:н — 2.3 Но:р %3D 0.575 Но: д —— 2.3 Но:р %3D 0.575 Но:р 3 0.575 H:р + 2.3 Hi:p 2.3 Н:р#0.575 H:р> 0.575 The test is: left-tailed right-tailed two-tailed Based on a sample of 65 people, the sample mean GPA was 2.34 with a standard deviation of 0.06 The test statistic is: (to 2 decimals) The critical value is: (to 2 decimals) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesisWhen asked what it means for the result of a hypothesis test to be practically significant (or practically important), Tatum says this means that the P-value is small and the null hypothesis has been rejected. what is wrong about this?
- Test the claim that the proportion of people who own cats is significantly different than 40% at the 0.2 significance level. The null and alternative hypothesis would be: Ho:p 0.4 Ho: u 2 0.4 Ho:p = 0.4 Ho: p = 0.4 Ho:u 0.4 H1:p 0.4 The test is: two-tailed left-tailed right-tailed Based on a sample of 700 people, 34% owned cats The p-value is: (to 2 decimals) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesis Question Help: DVideo M hpAgain, you play the game where Bianca rolls a die and wins if it shows a 1 or a 2, otherwise you win. Out of 7 games, how many times does she have to win at least so that you would reject the null hypothesis (that the die is fair, results random) at the 5%-significance level? You can look up numbers in the Binomial table, but remember that one number is not enough. You can also use the Online Calculator, but then you need to convert the answer to percent.WHen writing a null hypothesis for a two sample t test. Do you write U1 = U2 or U1 - U2 = 0? Or do both of them work?