When the independent variable truly does alter the dependent variable scores (there is a significant effect), and our calculated test statistic does not exceed the critical cutoff value, we retain the null hypothesis when we should reject it.  This kind of result is called? a.) a type 2 error b.) a type 1 error c.) a valid result d.) a large effect size

You may need to use the appropriate technology to answer this question. The following table contains observed frequencies for a sample of 200. RowVariable Column Variable A B C P 22 46 52 Q 28 24 28 Test for independence of the row and column variables using  ? = 0.05. State the null and alternative hypotheses. H0: Variable P is not independent of variable Q.Ha: Variable P is independent of variable Q.H0: The column variable is independent of the row variable.Ha: The column variable is not independent of the row variable.    H0: The column variable is not independent of the row variable.Ha: The column variable is independent of the row variable.H0: Variable P is independent of variable Q.Ha: Variable P is not independent of variable Q. Find the value of the test statistic. (Round your answer to three decimal places.)   Find the p-value. (Round your answer to four decimal places.) p-value =  State your conclusion. Reject H0. We conclude that there is an association…

Observations from The null hypothesis is Ho: H1 = H, and the alternative hypothesis is as specified. The provided data are from a simple random paired sample from the two populations under consideration. Use the paired t-test to perform the required hypothesis test at the 10% significance level. Pair Population 1 Population 2 24 20 Hai Hy> H2 2 9 10 20 19 E Click the icon to view the t-table. 4. 13 8 5 8 14 14 7 10 15 8 22 28 Find the test statistic. Use population 1- population 2 as the difference. t=| (Round to three decimal places as needed.)

is a simple random sample. 1) Ten students in a graduate program were randomly selected. Their grade point averages (GPAs) when they entering entered the program were between 3.5 and 4.0. The following data were obtained regarding their GPAs on program versus their current GPAs. the Entering GPA 3.5 3.8 3.6 3.6 3.5 3.9 4.0 3.9 3.5 3.7 Current GPA 3.6 3.7 3.9 3.6 3.9 3.8 3.7 3.9 3.8 4.0

Test the claim that the proportion of men who own cats is significantly different than 70% at the 0.1 significance level. The null and alternative hypothesis would be: Но: р 3 0.7 Но:д — 0.7 Но:д %3D 0.7 Но:р — 0.7 Но:р — 0.7 Но: д — 0.7 H:р +0.7 H:n # 0.7 Hi:р > 0.7 H:p > 0.7 H:р < 0.7 H:р < 0.7 The test is: two-tailed left-tailed right-tailed Based on a sample of 55 people, 75% owned cats The test statistic i: (to 2 decimals) The positive critical value is: (to 2 decimals) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesis

Test the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.1 significance level.The null and alternative hypothesis would be:   A.) H0:μM=μF     H1:μM≠μF   B.)H0:μM=μF    H1:μM<μF   C.)H0:pM=pF    H1:pM<pF   D.)H0:pM=pF     H1:pM≠pF   E.)H0:μM=μF    H1:μM>μF   F.)H0:pM=pF    H1:pM>pF The test is:   A.) right-tailed B.) left-tailed C.) two-tailed Based on a sample of 40 men, 40% owned catsBased on a sample of 20 women, 55% owned cats The test statistic is:___________  (to 2 decimals)The p-value is: _____________ (to 2 decimals)Based on this we: A.) Reject the null hypothesis B.) Fail to reject the null hypothesis

Many parents have turned to St. John's wort, an herbal remedy, to treat their children with attention deficit hyperactivity disorder (ADHD). The Journal of the American Medical Association (June 11, 2008) published an article that explored the herb's effectiveness. Children with ADHD were randomly assigned to take either St. John's wort capsules or placebos. The contingency table below broadly reflects the results found in the study. (You may find it useful to reference the appropriate table: chi-square table or F table) Effect on ADHD Treatment No Change in ADHD Improvement in ADHD St. John's wort 12 15 Placebo 14 13 At the 5% significance level, do the data indicate that St. John's wort affects children with ADHD? a. Choose the null and alternate hypotheses. O Ho: Effect on ADHD is independent of the treatment; HẠ: Effect on ADHD is dependent of the treatment O Ho: Effect on ADHD is dependent of the treatment; HA: Effect on ADHD is independent of the treatment b. Calculate the test…

Test the claim that the proportion of men who own cats is larger than 20% at the .05 significance level. The null and alternative hypothesis would be: 0.2 Но: р — 0.2 Но: д 0.2 Но:р Но: и Н: + 0.2 Н]:р 0.2 Hi:д > 0.2 Hi:р+0.2 0.2 Но:р — 0.2 Но: д 0.2 The test is: left-tailed two-tailed right-tailed Based on a sample of 25 people, 25% owned cats 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 Reject the null hypothesis

A nutritionist wants to investigate whether her new diet will be effective in helping women aged 30-40 to lose weight. She will use a paired sample to determine whether the mean weight of women before going on this diet is greater than the mean weight of women after being on this diet for two months. Determine the null and alternative hypotheses for the proposed hypothesis test. O A. Let μ₁ denote the mean weight of women before going on this diet and let μ₂ denote the mean weight of women who have been on this diet for two months. The null and alternative hypotheses are Ho: H₁ H2 and Ha: H₁ H₂. O D. Let μ₁ denote the mean weight of women before going on this diet and let μ₂ denote the mean weight of women who have been on this diet for two months. The null and alternative hypotheses are Ho: H₁ H2 and Ha: H₁ H₂.

Test the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.01 significance level. The null and alternative hypothesis would be: A.)H0:pM=pF          H1:pM>pF    B.)H0:pM=pF          H1:pM≠pF       C.)H0:μM=μF          H1:μM≠μF       D.)H0:pM=pF           H1:pM<pF        E.)H0:μM=μF          H1:μM>μF         F.)H0:μM=μFH0:μM=μF    H1:μM<μFH1:μM<μF The test is: A.) left-tailed B.) right-tailed C.) two-tailed Based on a sample of 60 men, 30% owned catsBased on a sample of 60 women, 50% owned cats The test statistic is:  (to 2 decimals) The positive critical value is:  (to 2 decimals) Based on this we: A.) Fail to reject the null hypothesis B.) Reject the null hypothesis