Question

A new kind of typhoid shot is being developed by a medical research team. The old typhoid shot was known to protect the population for a mean time of 36 months, with a standard deviation of 3 months. To test the time variability of the new shot, a random sample of 17 people were given the new shot. Regular blood tests showed that the sample standard deviation of protection times was 1.5 months. Using a 0.05 level of significance, test the claim that the new typhoid shot has a smaller variance of protection times. (a) What is the level of significance? State the null and alternate hypotheses. O Ho: 02 = 9; Hx: 029 OH:02 = 9; H7:02 <9 O He: 02 = 9; H7:02 + 9 OHO: 02 <9; H: 02 = 9 What are the degrees of freedom? What assumptions are you making about the original distribution? We assume a exponential population distribution. We assume a uniform population distribution. We assume a normal population distribution. We assume a binomial population distribution. (c) Find or estimate the p-value of the sample test statistic. OP-value > 0.100 O 0.050 < P-value < 0.100 O 0.025 < P-value < 0.050 O 0.010 < P-value < 0.025 O 0.005 < P-value < 0.010 OP-value < 0.005 (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Since the P-value > a, we fail to reject the null hypothesis. Since the P-value > a, we reject the null hypothesis. Since the P-value sa, we reject the null hypothesis. Since the P-value sa, we fail to reject the null hypothesis.

Answer 1

Given, n=17 Standard deviation, o=3 sample variance, S = 1.5.). vai noin 5² (115) ² op 'n 2.25 a) level of significance Son 10.05 - woll. Hypothesis: H: 6 - gelov ! Alternative Hypothesis: H obie ģ b) chi-square test Statistic, ish Toda Biudget. otestearg lo Danies (1740) (2.35) lorog ell - 01 16 (2.25)ai oubital 9 moisiwe Os = 36 19.00 2. Test Statistic, x'=14.00] → Degress of freedom, intito. In df sin 3.15-1 = [10 517-1

-> We assume a normal population distribution c) P-value P-value - p(x, dt) =p(4.00, 16) (using p-value from chi-square Score calculator) - 0.9989 P-value > 0.100 d) there, p-value > a ine, p-value 0.05 tot 10 Since the p-values &, we thypothesis. fail to reject the null e) At the 54. level of significance, there insufficient evidence to conclude that typhoid shot has a smalles variance lines. is the of new protection * 904. confidence interval a for the population deviation. Significance Level da 1-0.90 = 0.1 Riool = 0.05 Degrees of freedom, od fini (217) - 16

teende = x2005, 16) ( from chisquared table) = 7.962 = M-005, 16) = 40.95,16) : - 26.296 90.1. confidence interval is, (n-1)3 ((17-1) (2.25) (17-1) (2.25 sos- 7.962 26. 296 16(2-25) 1662- 25) sos 7.962 26.296 = 117 265 2.13 Lower limit = (117) upper limit = 2:13) we are goy. confident interval. that a lies within this

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