A researcher wants to determine if the proportion of adult learners who are full-time students differs from the proportion of traditional undergraduates who are full time. Nationally, 29% of undergraduates are full-time students; a survey of 215 adult learners found that 34% were full-time students. (a) State an appropriate null and alternate hypothesis symbolically. (b) Compute the observed value of the test statistic. Show the formula you used and the computations. (c) Calculate the p-value for the statistic. Show how you obtained this value. (d) Decide whether the hypothesis should be rejected or not at the .05 level and draw an appropriate conclusion in the context of the study. Clearly state the basis for your decision to reject or not. (e) How would the p-value have changed if a larger sample size had been used? Explain. (f) Construct a 95% confidence interval for the proportion of adult learners who are full-time learners. State in words what this interval means.
Added by Aaron C.
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H0: p1 = p2 Alternate hypothesis: The proportion of full-time students among adult learners is different from the proportion of full-time students among traditional undergraduates. Ha: p1 ≠ p2 Show more…
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Recently, a local newspaper reported that part-time students are older than full-time students. In order to test the validity of its statement, two independent samples of students were selected. Full-Time Part-Time (in years) 26 24 s 2 3 n 42 31 a. Give the hypotheses for the above. b. Determine the degrees of freedom. c. Compute the test statistic. d. Using α = .05, test to determine whether or not the average age of part-time students is significantly more than full-time students. ANSWER: a. H0: μp - μf ≤ 0 Ha: μp - μf > 0 b. 49 c. test statistic t = -3.221 d. p-value (.0011) is less than .005, reject H0; the average age of part-time students is significantly more. The following information is gathered from random samples of day and evening students regarding the number of semester hours they take. Day Evening 16 12 s 4 2 n 140 160 Develop a 95% confidence interval estimate for the difference between the mean semester hours taken by the two groups of students. ANSWER: 3.269 to 4.731 hours
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34. According to a nation wide survey, only 30% of adult Americans had a will. A researcher in Utah suspects that the proportion of adults in Utah with wills is less than the national proportion. Suppose this researcher takes a survey of 1000 adult Utah residents and finds that 277 report having wills. (round your answers to four decimal places) part a: Specify the null and alternative hypotheses to determine whether the proportion of adult Utah residents that have a will is less than the national proportion. part b: Calculate and report the value of the test statistic. part c: Calculate and report the p-value and then use the p-value approach to determine at a 10% level of significance if the proportion of adult Utah residents that have a will is less than the national proportion. part d: Determine and report the critical value(s) for a 10% level of significance and use the critical value approach to determine if the proportion of adult Utah residents that have a will is less than the national proportion.
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In a survey of 500 males ages 20 to 24, 15.8% were neither in school nor working. In a survey of 500 females ages 20 to 24, 17.8% were neither in school nor working. At α = 0.05, can you support the claim that the proportion of males ages 20 to 24 who were neither in school nor working is less than the proportion of females ages 20 to 24 who were neither in school nor working? (a) Identify the claim and state H0 and Ha, (b) Find the critical value(s) and identify the rejection region(s), (c) Find the standardized test statistic z, (d) Decide whether to reject or fail to reject the null hypothesis, and (e) Interpret the decision in the context of the original claim. Assume the samples are random and independent.
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