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Save Out H Homework: Section 7.5 Homework 13 of 15 (15 complete) HW Score: 86.27%, 12.94 of 15 pts Score: 0.86 of 1 pt Plus" 7.5.23 Question Help rses e Home A snack food manufacturer estimates that the variance of the number of grams of carbohydrates in servings of its tortilla chips is 1.24. A dietician is asked to test this claim and finds that a random sample of 16 servings has a variance of 1.08. At a=0.10, is there enough evidence to reject the manufacturer's claim? Assume the population is normally distributed. Complete parts (a) through (e) below. (a) Write the claim mathematically and identify Ho and H. bus dy Plan O A. Ho: 0221.24 Hy <1.24 (Claim) OC. Ho: 51.24 (Claim) H:02 > 124 O B. Ho: 02 1.24 H o2 = 1.24 (Claim) D. He: 62 = 1.24 (Claim) H:c?#1.24 radebook StatCrunch eText (b) Find the critical value(s) and identify the rejection region(s). The critical value(s) is(are) (Round to two decimal places as needed. Use a comma lo separate answers as needed.) Chapter Contents Accessible Resour Enter your answer in the answer box and then click Check Answer. Skills for Success Clear All Check Answer 3 parts remaining Tools for Success 4:40 PM 4/24/2020 Type here to search Lenovo

Answer 1

The provided sample variance is s2 = 1.08 and the sample size is given by n = 16. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: o2 = 1.24 Ha: 0% = 1.24 This corresponds to a two-tailed test, for which a Chi-Square test for one population variance will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.10, and the the rejection region for this two-tailed test is R= {x? : x? < 7.261 or x? > 24.996} (3) Test Statistics The Chi-Squared statistic is computed as follows: 12_(n − 1)s? (16 - 1). 1.08 x = 2 = 13.065 1 .24 (4) Decision about the null hypothesis Since it is observed that xź = 7.261 < x² = 13.065 < xi = 24.996, it is then concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population variance o is different than 1.24, at the 0.10 significance level. Confidence Interval The 90% confidence interval is 0.6481 < 0º < 2.2311.

Critical values are lower = 7.261 and upper = 24.996

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