Please help me with these. Thank you. Please answer it like the examples.

\f1. In a time motion study, it was found out that the time required by workers to complete a certain manual operation was normally distributed with a mean of 26.6 minutes with a standard deviation of 3 minutes. A group of 25 workers was randomly selected to receive a special training for two weeks. After the training, it was found that their average time to finish the operation was 23 minutes. Can it be concluded that the special training speeds up the operation? Use 0.05 level of significance.2. An insurance company claims that the mean cost to process a claim is $60. An industry comparison showed this amount to be larger than most other insurance companies, so the company instituted cost-cutting measures. To evaluate the effect of the cost-cutting measures, the supervisor selected a random sample of 26 claims processed last month and recorded the cost to process each claim. The sample information is reported below (in $). 45 49 62 40 43 61 48 53 67 63 78 64 48 54 51 56 63 69 58 51 58 59 56 57 38 76 At the 0.05 significance level, is it reasonable to conclude that the mean cost to process a claim is now less than $60?3. A new papaya soap manufacturer claims that 80% of women who used the soap observed skin whitening within two weeks. One of their competitors makes a random survey if the claim is true. The survey results in 6 out of 10 women observed skin whitening within two weeks. With a=5%, is the soap manufacturer's claim true?' Example 1: In a study, it was found out that the average IQ score of students is 101.5. The variable is normally distributed, and the population standard deviation is 15. A school superintendent claims that the students in her school district have an IQ higher than the average of 101.5. She selects a random sample of 30 students and finds the mean of the test scores is 106.4. Test the claim at a = 0.05. Solution: Step 1: State the null (Ho) and alternative (Ha) hypothesis. Identify the claim. Ho : ILt = 101.5 => The average I0. score of students is 101.5 Ha : ,u > 101.5 => The average IQ score of students is higher than 101.5 {claim} Step 2: Determine the test statistic, critical value and tail of the distribution where the rejection region is located. The test to use is z-test since the data is normally distributed and the population standard deviation cr is known. Since the alternative hypothesis is lit 2* 101.5, therefore the tail of the distribution is right-tailed test. ".__ Rejection Reglor U z = +1.645 Since a: = 0.05 and the test is a right-tailed test, the critical value is Z\" = 20.05 = +1. 64-5. Step 3: Formulate the decision rule. Since the alternative hypothesis is 11 "9- 101.5, therefore the decision rule is to Reject Ha if Zc :- zoos = 1. 645. Step 4: Compute the value of the test statistic. Given:,u = 101.5, 0' = 15, n = 30, X = 106.4- X a 106.4 101.5 2'. = == 1.79 J/J 15/1!E STATISTICS Step 5: Make a decision. Since the value of test statistic is 1.79 which is greater than the critical value 1.645, the decision is to reject the null hypothesis. Rejection Region Acceptance Region O 1.645 1.79 Step 6: Draw conclusion There is enough evidence to support the claim that the IQ of the students is higher than the state average