The null hypothesis and the alternate are H0: The frequencies are equal. H1: The frequencies are not equal. Category f0 A 15 B 10 C 10 D 15
Solution:
The null hypothesis and the alternate are
H0: The frequencies are equal.
H1: The frequencies are not equal.
The test statistic is:
The null hypothesis and the alternate are H0: The frequencies are equal. H1: The frequencies are...
The null hypothesis and the alternate hypothesis are: H0: The frequencies are equal. H1: The frequencies are not equal. Category f0 A 25 B 20 C 25 D 10 1) State the decision rule, using the 0.01 significance level. Reject H0 if chi-square > _____ 2) Compute the value of chi-square.
The null hypothesis and the alternate hypothesis are: Ho: The frequencies are equal. Hi: The frequencies are not equal. Category A B с fo 10 20 30 20 D a. State the decision rule, using the 0.05 significance level. (Round your answer to 3 decimal places.) Reject HO if chi-square > b. Compute the value of chi-square. Chi-square value c. What is your decision regarding Ho? HO. The frequencies are
The null and alternate hypotheses are: H0 : μ1 = μ2 H1 : μ1 ≠ μ2 A random sample of 11 observations from Population 1 revealed a sample mean of 21 and sample deviation of 3.5. A random sample of 7 observations from Population 2 revealed a sample mean of 23 and sample standard deviation of 3.8. The underlying population standard deviations are unknown but are assumed to be equal. At the .05 significance level, is there a...
The null and alternate hypotheses are: H0 : μ1 = μ2 H1 : μ1 ≠ μ2 A random sample of 8 observations from Population 1 revealed a sample mean of 25 and sample deviation of 4.5. A random sample of 8 observations from Population 2 revealed a sample mean of 26 and sample standard deviation of 3.5. The underlying population standard deviations are unknown but are assumed to be equal. At the .05 significance level, is there a difference between...
The null and alternate hypotheses are: H0: μ1 ≤ μ2 H1: μ1 > μ2 A random sample of 27 items from the first population showed a mean of 110 and a standard deviation of 15. A sample of 19 items for the second population showed a mean of 100 and a standard deviation of 6. Use the 0.025 significant level. a. Find the degrees of freedom for unequal variance test. (Round down your answer to the nearest whole number.) b....
The null and alternate hypotheses are: H0 : μd ≤ 0 H1 : μd > 0 The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month. Day 1 2 3 4 Day shift 12 12 16 19 Afternoon shift 10 10 12 15 At the .05 significance level, can we conclude there are more...
When 50 people used the Weight Watchers diet for one year, their mean weight loss was 3.0 lb and standard deviation was 4.9lb. Assume that the population standard deviation is 5.0 lb. Use a 0.01 significance level to test the claim that the mean weight loss is greater than 1.5.Find the null hypothesis and alternate hypothesis.H0: [ Select ] ["p", "mu"] ...
The null and alternate hypotheses are: H0 : μd ≤ 0 H1 : μd > 0 The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month. Day 1 2 3 4 Day shift 11 12 14 18 Afternoon shift 9 10 13 16 At the .005 significance level, can we conclude there are more defects produced on the afternoon shift? Hint: For the...
The null and alternate hypotheses are: H0 : μd ≤ 0 H1 : μd > 0 The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month. Day 1 2 3 4 Day shift 10 12 15 19 Afternoon shift 8 11 12 20 At the 0.050 significance level, can we conclude there are more defects produced on the day shift? Hint: For the...
The null and alternate hypotheses are: H0 : μd ≤ 0 H1 : μd > 0 The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month. Day 1 2 3 4 Day shift 10 10 16 17 Afternoon shift 9 10 14 15 At the 0.100 significance level, can we conclude there are more defects produced on the day shift? Hint: For the...