The table below gives flight arrival numbers from a random sample of flights for two airlines. Early On-time 355 416 771 Late Total Airline A 87 58 133 191 500 151 700 Airline B Total 238 1200 3.1 Determine whether the proportion of on-time flights from Airline B exceeds that of Airline A by using the appropriate hypothesis test. Show all 7 steps and include both the rejection region and the p-value. (9) 3.2 By how much does the proportion of on-time flights for Airline B exceed that of Airline A? Use a 95% lower confidence bound to answer this question. Show all workings. (5)
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- In a recent study of a random sample of 113 people, 92 reported dreaming in color. In the 1940’s, the proportion of people who reported dreaming in color was 29%. Is there evidence to show that the proportion of people who dream in color is different than the proportion in the 1940’s? What is the critical value for this hypothesis test? In a recent study of a random sample of 113 people, 92 reported dreaming in color. In the 1940’s, the proportion of people who reported dreaming in color was 29%. Is there evidence to show that the proportion of people who dream in color is different than the proportion in the 1940’s? What is the critical value for this hypothesis test? \pm1.96±1.96 \pm1.645±1.645 \pm2.575±2.575 \pm2.05±2.05 \pm2.33±2.33A well-known brokerage firm executive claimed that 60% of investors are currently confident of meeting their investment goals. An XYZ Investor Optimism Survey, conducted over a two week period, found that in a sample of 300 people, 65% of them said they are confident of meeting their goals.Test the claim that the proportion of people who are confident is larger than 60% at the 0.01 significance level.The null and alternative hypothesis would be: H0:μ≥0.6H0:μ≥0.6H1:μ<0.6H1:μ<0.6 H0:μ=0.6H0:μ=0.6H1:μ≠0.6H1:μ≠0.6 H0:p=0.6H0:p=0.6H1:p≠0.6H1:p≠0.6 H0:p≥0.6H0:p≥0.6H1:p<0.6H1:p<0.6 H0:μ≤0.6H0:μ≤0.6H1:μ>0.6H1:μ>0.6 H0:p≤0.6H0:p≤0.6H1:p>0.6H1:p>0.6 The test is: left-tailed two-tailed right-tailed The test statistic is: (to 3 decimals)The p-value is: (to 4 decimals)Based on this we: Reject the null hypothesis Fail to reject the null hypothesisTo determine whether training in a series of workshops on creative thinking increases IQ scores, a total of 70 students are randomly divided into treatment and control groups of 35 each. After two months of training, the sample mean IQ for the treatment group equals 110, and the sample mean IQ for the control group equals 108. The estimated standard error equals 1.80. Using t, test the null hypothesis at the .01 level of significance. If appropriate (because the null hypothesis has been rejected), estimate the standardized effect size, construct a 99 percent confidence interval for the true population mean difference, and interpret these estimates.