7 **Hypothesis Testing and Confidence Interval Construction for IQ Scores Based on Lead Levels**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.**IQ Scores and Data:**- Medium Lead Level IQ Scores: 72, 96, 92, 85, 88, 97, 83, 92, 96, 111, 91- High Lead Level Statistics: - Sample Size (\(n_2\)) = 11 - Mean (\(\bar{x}_2\)) = 89.723 - Standard Deviation (\(s_2\)) = 9.629---**a. Test Using a 0.05 Significance Level****Objective:** 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.**Hypotheses:**- **A.** \(H_0: \mu_1 = \mu_2\); \(H_1: \mu_1 > \mu_2\)- **B.** \(H_0: \mu_1 \neq \mu_2\); \(H_1: \mu_1 > \mu_2\)- **C.** \(H_0: \mu_1 \le \mu_2\); \(H_1: \mu_1 > \mu_2\)- **D.** \(H_0: \mu_1 = \mu_2\); \(H_1: \mu_1 > \mu_2\)**Results:**- **Test Statistic:** [Fill in and round to two decimal places]- **P-value:** [Fill in and round to three decimal places]**Conclusions:**- **A.** Reject the null hypothesis. There is not sufficient evidence to support the claim that subjects with medium lead levels have higher IQ scores.- **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

The data shows the IQ scores for a random sample of subjects with medium lead levels and high lead levels in their blood.a. Let μ1 be the population mean IQ scores for subjects with medium lead levels, μ2 be the population mean IQ scores for subjects with high lead levels. The claim is that the mean IQ scores for subjects with medium lead levels is higher than the mean for subjects with high lead level. The hypothesis is, Null hypothesis: H0:μ1=μ2 Alternative hypothesis: H1:μ1>μ2 Correct Answer: Option D. The mean for medium lead level is, x¯1=∑x1n1=72+96+92+85+88+97+83+92+96+111+9111=100311=91.1818 The sample variance for medium lead level is, s12=∑x1-x¯12n1-1=72-91.18182+96-91.18182+92-91.18182+⋯+91-91.1818211-1=95.7636 The test statistic is, t=x¯1-x¯2s12n1+s22n2=91.1818-89.72395.763611+9.629211=1.45884.13940=0.3524≈0.35 Therefore, the test statistic is 0.35. The degrees of freedom is, df=s12n1+s22n22s12n12n1-1+s22n22n2-1=95.763611+9.629211295.763611211-1+9.629211211-1≈19 The hypothesis is right tail. The probability of t greater than 0.3524 with 19 degrees of freedom can be obtained using the excel formula “=T.DIST.RT (0.3524, 19)”.The probability value is 0.3642. Therefore, the P-value is 0.364. Decision rule: If P-value≤α, then reject the null hypothesis. Conclusion: The P-value (0.364) is greater than the level of significance (0.05). Based on decision rule, fail to reject the null hypothesis. Thus, there is not sufficient evidence to support the claim that subjects with medium lead levels have higher IQ scores. Correct Answer: Option B.b. Since, the test is one tailed with 0.05 level of significance, the two tailed level of significance should be 0.10. Computation of critical value: The critical value of t-distribution for 0.10 level of significance at 19 degrees of freedom can be obtained using the excel formula “=T.INV.2T (0.10, 19)”. The critical value is 1.7291. The confidence interval is, x¯1-x¯2-tcs12n1+s22n2<μ1-μ2<x¯1-x¯2+tcs12n1+s22n291.1818-89.723-1.729195.763611+9.629211<μ1-μ2<91.1818-89.723+1.729195.763611+9.6292111.4588-7.1574<μ1-μ2<1.4588+7.1574-5.70<μ1-μ2<8.62 Therefore, the confidence interval is -5.70<μ1-μ2<8.62. Yes, because the confidence interval contains zero.

Math

Statistics

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. E Click the icon to view the data table of IQ scores. C a. Use a 0.05 significance level to test the claim that the mean IQ soores 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. OA. Ho: H1 "P2 OB. Ho: HH2 OC. Ho: H SP OD. Ho: "P2 H: H P2 IQ scores The test statistic is (Round to two decimal places as needed.) Medium Lead Level o High Lead Level The P-value is . (Round to three decimal places as needed.) 72 96 92 85 88 97 83 ng11 State the conclusion for the test. X- 89.723 OA. Reject the null hypothesis. There is not 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 evidenoe to support the claim that subjects with medium lead levels have higher IQ scores. OC. Reject the null hypothesis. There is sufficient evidence to support the claim that subjects with medium lead levels have higher IQ scores. OD. Fail to reject the null hypothesis. There is sufficient evidence to support the claim that subjects with medium lead levels have higher IQ scores. - 9.629 92 96 111 91 b. Construct a confidence interval suitable for testing the claim that the mean IQ scores for subjects with medium lead levels is higher than the mean for subjects with high lead levels. (Round to two decimal places as needed.) Print Done Does the confidence interval support the conclusion of the test? V because the confidence interval contains