An investigator thinks that people under the age of forty have vocabularies that are different than those of people over sixty years of age. The investigator administers a vocabulary test to a group of 37younger subjects and to a group of 45older subjects. Higher scores reflect better performance. The mean score for younger subjects was 15and the standard deviation of younger subject's scores was 5. The mean score for older subjects was 20and the standard deviation of older subject's scores was 5.9. Does this experiment provide evidence for the investigator's theory?Useα=0.10) a. Write alternative. Do not write null hypothesis. Indicate the tail. b. Find test statistics (round to2decimals) c. Find P-value (round to 3decimal places). Can we reject the null hypothesis? d. Write the conclusion in the contest of the problem e. Find a 90% confidence interval for μ1–μ2 (round to2decimals)
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Step 1: Set up the alternative hypothesis: \(H_1: \mu_1 \neq \mu_2\) (two-tailed test) Show more…
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A store owner surveyed 25 randomly selected customers and found the ages shown (in years). The mean is 29.84 and the standard deviation is 10.74. The owner wants to know if the mean age of all customers is 34 years old. Use the given information to complete parts a through f. 34 41 30 27 34 36 48 19 36 32 19 48 27 41 16 36 36 15 18 23 11 28 13 41 37 H0: μ = 34 (Type an integer or a decimal.) b) Is the alternative one or two-sided? The alternative hypothesis is two-sided. c) What is the value of the test statistic? The test statistic is . (Round to two decimal places as needed.) d) What is the P-value of the test statistic? P-value = (Round to four decimal places as needed.) e) What do you conclude at α = 0.10? The p-value is α, so the null hypothesis. There is evidence to conclude that the mean age of all customers is not 34 years old.
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