O c Chegg Study l Guided Solu x M Tests (P Take a Test Zachary hicks x C https:
ID: 3221452 • Letter: O
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O c Chegg Study l Guided Solu x M Tests (P Take a Test Zachary hicks x C https: 151701460¢erwin; yes Apps Watch Kids Movies NHL NETWORK Li D Ce age Learning D Credit Card Acco G Computershare i... e SEC.gov l compan... D HD Sports Watch Other Bookmarks 172BUS207.920 Business Statistics Spring 2017 zachary hicks I 4/13/17 11:23 PM Test Ch. 10 Test Time Limit: 02:00:00 Submit Test This Question: 1 pt 7 of 12 (0 complete) This Test: 12 pts possible Question Help The following table shows the results of two random samples that measured the average number of minutes per charge for AA Lithium-ion (Li-ion) rechargeable batteries versus Nickel-Metal Hydride (NiMH) rechargeable batteries. Complete parts a through c. Li ion NiMH Sample mean 96.4 82.8 Sample standard deviation Sample size a. Perform a hypothesis test using ot 0.01 to determine if the average number of minutes per charge differs between these two battery types. Assume the population variances for the number of minutes per charge are not equal Determine the null and alternative hypotheses for the test. Ho: u H2 Calculate the appropriate test statistic and interpret the result. The test statistic is (Round to two decimal places as needed.) The critical values) is(are) (Round to two decimal places as needed. Use a comma to separate answers as needed.) the null hypothesis. Because the test statistic b. Identify the p-value from part a and interpret the result. The p-value is (Round to three decimal places as needed,) Interpret the result. Choose the correct answer below. O A. Since the p-value is less than the significance level, reject the null hypothesis. Click to select your answer(s).Explanation / Answer
The statistical software output for this problem is:
Two sample T hypothesis test:
1 : Mean of Population 1
2 : Mean of Population 2
1 - 2 : Difference between two means
H0 : 1 - 2 = 0
HA : 1 - 2 0
(without pooled variances)
Hypothesis test results:
So,
H0 : 1 - 2 = 0
HA : 1 - 2 0
Test statistic = 4.27
Critical values for 25.624 degrees of freedom and 0.01 level of significance = -2.78, 2.78
Because the test statistic is in the rejection region, reject the null hypothesis.
b) p - value = 0.000
Option A is correct.
Difference Sample Diff. Std. Err. DF T-Stat P-value 1 - 2 13.6 3.1870325 25.62419 4.2672925 0.0002Related Questions
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