Question

Listed in the data table are IQ scores for a random sample of subjects with medium lead levels in their blood. Also listed are statistics from a study done of IQ scores for a random sample of subjects with high lead levels. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts. Click the icon to view the data table of IQ scores a. Test the claim that the mean IQ scores for subjects with medium lead levels is higher than the mean for subjects with high lead levels What are the null and alternative hypotheses? Assume that population 1 consists of subjects with medium lead levels and population 2 consists of subjects with high lead levels. OB. Ho-11 =12 H:17 >12 IQ Scores x O A. Ho HH2 Hp : H1>H2 O c. Ho:Hy H2 H:H1>H2 OD. HOH = 12 H: H+H2 The test statistic is (Round to two decimal places as needed.) The P-value is (Round to three decimal places as needed.) Medium Lead Level High Lead Level 72 n = 11 81 92 X2 = 89.280 85 85 S2 = 10.613 97 83 92 102 111 91 State the conclusion for the test. O A. Reject the null hypothesis. There is sufficient evidence to support the claim that subjects with medium lead levels have higher IQ scores. O B. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that subjects with medium lead levels have higher IQ scores. O C. Fail to reject the null hypothesis. There is sufficient evidence to support the claim that subjects with medium lead levels have higher IQ scores. OD Reiert the null hypothesis. There is not sufficient evidence to support the claim that subiects with medium lead levels have higher 10 scores Print Done Click to select your answer(s).

Answer 1

9) Ho: M = M₂ Ha i su option (c) sie du b ) SY,: 991 EU 99111 11 90.09091 standard deviation sa { (2-2) ? nt (72-9010909097 7791-96.0909) 11-1 110.1093 = 89.280 2 i 90-09091 Sz. 10-613 10-7093 ne ... 1) an

7-ke 6: SY ni SZ 90.09091- 99.900 (10.7093) (16.613) ū 0-81091 4.546 0.1783 Sh รา SE nz + df: 2 SI S 7 na nil 2 10-70932 10613 ( 10-70932 10:2093% (105613 t 10 10 19.9984 » 20.

P-value: 0.4301 Prvalue (0.4301) > X10:05) thene in not Fail to reject the null hypothesis claim that Sufficient Evidence to support the lead level have medium with m Subjects highest sa scores option ni-B