When plotting these samples in stat disk my answer is matching up

with a tutor document I found online, but the confidence interval numbers are different. Would you create a sample and see if I’m right or wrong? The statdisk is under Elementary Statistics 11 th edition Passive and Active smoker. # 14 step 3
Alternative Hypothesis: from course hero
µ not equal to µ(hyp)

t Test
Test Statistic, t: 2.7745
Critical t: ±2.0227
P-Value: 0.0084

95% Confidence interval:
16.41367 < µ < 104.7363
Question 14 and 15
Alternative Hypothesis: from when I enter values
µ not equal to µ(hyp)

t Test
Test Statistic, t: 2.7745
Critical t: ±2.0227
P-Value: 0.0084

95% Confidence interval:
16.41364 < µ < 104.7364

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Part IV. Hypothesis Testing14.Your researcher’s claim is that the continine level for the ETS group is not equal to zero.Compose a hypothesis test with a level of significance of .05 to test your claim.Step 1.Determine parameter ofinterest and compose null andalternative hypotheses.The parameter of interest is the population mean contininelevel for the ETS group. As we want to test whether it is notequal to zero or not so the hypotheses are,H0:µ=0∧Ha:µ≠0Step 2.Determine the samplemean, sample standard deviation,and sample size.[Hint:Yourecorded these previously in PartII, #3-6]Sample Size, n: 40Mean:60.575St. Dev., s:138.0839Step 3.Determine the likelihoodthat the population mean isactually not equal to 0 bycompleting a Hypothesis Test:One Meanin STAT DISK.Usesignificance of0.05.Rememberto change the pulldown option inStat Disk to agree with thealternate hypothesis. Paste yourresults here.Alternative Hypothesis:µ not equal to µ(hyp)t TestTest Statistic, t: 2.7745Critical t:±2.0227P-Value:0.008495% Confidence interval:16.41367 <µ < 104.7363Step 4.State your conclusion.Your conclusion statementshould include the p-valueand level of significance tophrase your conclusion.The p-value is smaller than the significance level of 0.05thus we are rejecting the null hypothesis at 0.05significance level. The conclusion is that the populationmean is significantly different than 0 at 0.05 significancelevel.15. For your second hypothesis test, your researcher’s claim is that the continine level in the NOETS group is greater than zero.Perform a hypothesis test using a significance level of .01 to testyour claim.Step 1.Determine parameter ofinterest and compose null andalternative hypotheses.The parameter of interest is the population mean contininelevel for the NO ETS group. As we want to test whether it isgreater than zero or not so the hypotheses are,H0:µ≤0∧Ha:µ>0Step 2.Determine the samplemean, sample standard deviation,and sample size.[Hint:Yourecorded these previously in PartII, #3-6]Sample Size, n: 40Mean:16.35St. Dev., s:62.5335

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Step 3.Determine the likelihoodthat the population mean isactually greater than 0 bycompleting a Hypothesis Test:One Meanin STAT DISK.Usesignificance of0.01.Rememberto change the pulldown option inStat Disk to agree with thealternate hypothesis. Paste resultshere.Alternative Hypothesis:µ >µ(hyp)t TestTest Statistic, t: 1.6536Critical t:2.4258P-Value:0.053198% Confidence interval:-7.635266 <µ < 40.33527Step 4.State your conclusion.Your conclusion statementshould include the p-valueand level of significance tophrase your conclusion.The p-value is not smaller than the significance level of 0.01thus we are not rejecting the null hypothesis at 0.01significance level. The conclusion is that the populationmean is not significantly larger than 0 at 0.01significance level.16. Answer the following questions based on the above hypothesis test.a.What is the p-value forthe hypothesis test in #15and what does itrepresent? (Look at page383 in the text)The p-value is 0.0531. It represents that, if the nullhypotheses is true then the probability of obtaining thetest statistic which we obtained i.e. 1.6536 or larger is0.0531.b.Given that your data,hypotheses, and p-valuedo not change,whatwould need to bedifferent in order for youto REJECT the nullhypothesis? (What doyou compare to makeyour rejection decision?)The significance level should be higher than the p-value of0.0531.