The first pic is the question. Since the tutors are answering incorrectly more frequently, there’s a picture of an “example” to solving the problem. Sample QuestionCigarette Smoking A researcher found that a cigarettesmoker smokes on average 32 cigarettes a day. She feelsthat this average is too high. She selected a random sampleof 9 smokers and found that the mean number of cigarettesthey smoked per day was 29. The sample standard deviationwas 2.9. At a-0.01, is there enough evidence to support herclaim? Assume that the population is approximatelynormally distributed. Use the critical value method andtables.(2) State the hypotheses and identify the claim.(b) Find the critical value.(e) Compute the test value.(d) Make the decision.(e) Summarize the results.Explanation(a) State the hypotheses and identify the claim.The null hypothesis n, is the statement that there is nodifference between a parameter and a specific value. This isequivalent to H:- 32 cigarettes.The alternative hypothesis I, is the statement that there isa difference between a parameter and a specific value. Inthis case, the mean number of cigarettes smoked per day isless than 32. This is equivalent to il, :

Question

The first pic is the question. Since the tutors are answering incorrectly more frequently, there’s a picture of an “example” to solving the problem. Sample QuestionCigarette Smoking A researcher found that a cigarettesmoker smokes on average 32 cigarettes a day. She feelsthat this average is too high. She selected a random sampleof 9 smokers and found that the mean number of cigarettesthey smoked per day was 29. The sample standard deviationwas 2.9. At a-0.01, is there enough evidence to support herclaim? Assume that the population is approximatelynormally distributed. Use the critical value method andtables.(2) State the hypotheses and identify the claim.(b) Find the critical value.(e) Compute the test value.(d) Make the decision.(e) Summarize the results.Explanation(a) State the hypotheses and identify the claim.The null hypothesis n, is the statement that there is nodifference between a parameter and a specific value. This isequivalent to H:- 32 cigarettes.The alternative hypothesis I, is the statement that there isa difference between a parameter and a specific value. Inthis case, the mean number of cigarettes smoked per day isless than 32. This is equivalent to il, :<32 cigarettes.The problem asks, &#34;Is the mean number of cigarettessmoked per day less than 32;&#34;Hence, the claim is the alternative hypothesis.(b) Find the critical value.From OThe t Distribution Table, for a left-tailed test with a- 0.01 andd.f. -9-1-8, the critical value is --2.896.TABLE F The DistributionConfidenceintervalsOne tail, al80%90%95%98%99%d.f.0.100.050.0250.010.005Two tails0,200,100.050.020.011,4151,8952.3652.9983.49981.3971.8602.3062.8963.35591.3831.8332.2622.8213.250(c) Compute the test value.The test is a statistical test for the mean of apopulation and is used when the population isnormally or approximately normally distributed andis unknown.The formula for the test isThe degrees of freedom are d.f. -n-1Using the formula for the r-test,29 - 322.9//9--3.103Hence, the test value, rounded to 3 decimal places, isI--3.103.(d) Make the decision.Since the test value does fall in the critical region, reject thenull hypothesis.-3.103 -2.896(e) Summarize the results.Sinco tho pull bynothocic waC roioctod thoro ic onOugh Cigarette Smoking A researcher found that a cigarette smoker smokes on average 31 cigarettes a day. She feels that thisaverage is too high. She selected a random sample of 10 smokers and found that the mean number of cigarettes they smoked perday was 28. The sample standard deviation was 2.7. At a = 0.05, is there enough evidence to support her claim? Assume that thepopulation is approximately normally distributed. Use the critical value method and tables.Part: 0 /5Part 1 of 5(a) State the hypotheses and identify the claim.Ho(Choose one) ▼H(Choose one) ▼This hypothesis test is a (Choose one) ▼test.

Accepted Answer

The random variable X follows normal distribution.
We have to test whether the mean is less than 31…