Consider a paintdrying situation in which drying time for a test specimen is normally distributed with a = 8. The hypotheses Ho: p = 75 and Ha: u

I missed the first part of this problem and I cannot figure out why my part a is wrong if you could help me. The problem round to two decimals

—Consider a paint—drying situation in which drying time for a test specimen is normally distributed with a = 8. The hypotheses Ho: p = 75 and Ha: u < 75 are to be tested using a random sample of n = 25 observations. (a) How many standard deviations (of?) below the null value is )7 = 72.3? (Round your answer to two decimal places.) -1.69 X standard deviations (b) If)? = 72.3, what is the conclusion using a = 0.005?Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)2 = P-value = State the conclusion in the problem context.0 Do not reject the null hypothesis. There is sufficient evidence to conclude that the mean drying time is less than 75. Q Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean drying time is less than 75.Q Reject the null hypothesis. There is not sufficient evidence to conclude that the mean drying time is less than 75. Q Reject the null hypothesis. There is sufficient evidence to conclude that the mean drying time is less than 75. (c) For the test procedure with a = 0.005, what is fl(70)? (Round your answer to four decimal places.) 500) = i (d) If the test procedure with a = 0.005 is used, what 17 is necessary to ensure that [3(70) = 0.01? (Round your answer up to the next whole number.) n = specimens (e) If a level 0.01 test is used with n = 100, what is the probability of a type I error when p = 76.7 (Round your answer to four decimal places.) You may need to use the appropriate table in the Appendix of Tables to answer this question. Need Help? #—