1) Assume that the sample is an SRS. The problem state that the subjects were randomly selected.

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1 12.1 Homework for t Hypothei Tet 1) Below are the etimate of the daily intake of calcium in milligram for 38 randomly elected women between the age of 18 and 24 year who agreed to participate in a tudy of women bone health a) Contruct a 99% confidence interval for the true mean daily calcium intake for women age State: We want to etimate μ, the true mean daily calcium intake for all women between the age of 18 to 24. Plan: One-ample t confidence interval (σ unknown) with a 99% Confidence level 1) Aume that the ample i an SRS. The problem tate that the ubject were randomly elected. 2) The boxplot i relatively ymmetric. There i a light kew, but with no outlier and a ample ize of 38, we can aume that the ampling ditribution i approximately Normal. 3) Becaue the reearcher ample without replacement, aume that there are at leat 10(38) = 380 women between the age of 18 to 24 in the population. Alo aume that the calcium intake of each woman in the tudy wa independent of the other. Degree of freedom: 37 CI = x ± t = CI = x ± t = ± (2.715) = (741.17, ) We are 99% confident that the true mean daily calcium intake for all women between the age of 18 to 24 i between and milligram. b) A nutritionit believe that the average calcium intake for women age 18 to 24 i 739 milligram per day. If we were to tet thi claim baed on the confidence interval above, tate the null and alternative hypothei and the alpha level, then tate your concluion. H0: μ = 745 HA: μ 745 Alpha = 0.01 Becaue the null hypotheized value of 739 milligram per day i not included within our interval, we can reject the null hypothei at the 1% level. We do have evidence that the average calcium intake i not 739 milligram (we can refute the nutritionit claim). 2) The compoition of the earth atmophere may have changed over time. One attempt to dicover the nature of the atmophere long ago tudie the ga trapped in bubble inide ancient amber. Amber i tree rein that ha hardened and been trapped in rock. The ga in bubble within amber hould be a ample of the atmophere at the time the amber wa formed. Meaurement on pecimen of amber from the late Cretaceou era (75 to 95 million year ago) give thee percent of nitrogen: Are thee value ignificantly le than the preent 78.1% of nitrogen in the atmophere? Aume (thi i not yet agreed on by expert) that thee obervation are an SRS from the Cretaceou atmophere. State: μ: the true mean percent of nitrogen in the earth atmophere during the Cretaceou era. H0: μ = 78.1 HA: μ < 78.1 Plan: One-ample t tet (σ unknown)

2 1) Aume that the ample i an SRS. (Stated in the problem) 2) The boxplot of the ample data i kewed. With a ample ize of only 9, thi kew may indicate a non-normal the ampling ditribution. The normality condition ha not been met, o we will proceed with caution. 3) Aume that the meaurement taken were independent of one another. Degree of freedom = 8 = = p-value= There i a % change of getting a ample mean a extreme a 59.59% if the population mean i 78.1%. Becaue thi i VERY unlikely to occur, we can reject our null hypothei. We have good evidence that the true mean percent of nitrogen in the earth atmophere during the Cretaceou era i le than 78.1%. 3.) White blood cell count are normally ditributed with mean If a patient ha taken 50 laboratory blood tet that have a mean of and a tandard deviation of , doe thi give evidence at the 1% level that hi white blood cell count i ignificantly different than normal? State: We want to etimate μ, the true mean white blood cell count for all patient. H0: μ = 7500 HA: μ 7500 Plan: One-ample t tet (σ unknown) 1) Aume that the ample i an SRS. The problem did not tate anything regarding ampling method. 2) We are told the population of white blood cell count i normally ditributed. 3) Aume that the individual meaurement were independent of one another = = df = 49 Concluion: Becaue our p-value i le than our ignificance level of 0.01, we can reject H0. There i evidence that the true mean white blood cell count for all patient i not ) For each of the following, decide if it decribe 1 ample, 2 independent ample, or 2 dependent ample? a) We are teting to ee if the mean volume of a bag of regular m&m i equal to the tated volume of 8 ounce. ONE SAMPLE

3 b) We are teting to ee if the mean volume of a bag of regular m&m i equal to the mean volume of a bag of peanut m&m. TWO INDEPENDENT SAMPLES c) We are teting to ee if there i a preference of regular m&m or generic chocolate candie by doing a blind tate tet in which ubject eat both kind of chocolate candie in a randomly elected order, and rank both on a cale from DEPENDENT SAMPLES (matched pair) 5) An experiment wa done by 15 tudent in a tatitic cla at the Univerity of California at Davi to ee if manual dexterity wa better for the dominant hand compared to the nondominant hand (left or right). Each tudent meaured the number of bean they could place in a cup in 15 econd, once with the dominant hand and once with the nondominant hand. The order in which the two hand were meaured wa randomized for each tudent. Student Dominant hand Nondominant hand Difference a) Explain why the order of the two hand wa randomized rather than, for intance, having each tudent tet the dominant hand firt. If the order wa not randomized, the order could be a confounding variable. For example, if every tudent did the tet with their non-dominant hand firt, the practice/experience could explain why they did better when they ued their dominant hand. b) Compute a 90% confidence interval for the mean difference in the number of bean that can be placed into a cup in 15 econd by the dominant and nondominant hand. FULL PROCESS! State: μ: the true mean difference in number of bean that can be placed into a cup in 15 econd (dominant hand nondominant hand) Plan: One ample t confidence interval (σ unknown) 1) Aume that the ample i an SRS. The problem did not tate anything regarding ampling method. The order of the two hand wa randomized. 2) The boxplot how a light kew, but with a ample ize of 15 we can ay that the normality condition had been met. 3) Aume that the individual manual dexterity meaurement were independent of one another. Becaue the reearcher ampled without replacement, aume that there are at leat 10(15) = 150 tudent in the population. Degree of freedom: 14 CI = x ± t = CI = ± t = ( , ) We are 90% confident that the true mean difference in number of bean that can be placed into a cup in 15 econd (dominant hand nondominant hand) i between and

4 c) Ue the interval to addre the quetion of whether manual dexterity i better, on average, for the dominant hand. H0: μ diff = 0 Becaue 0 i included in the interval, we cannot reject our null hypothei. We do HA: μ diff > 0 not have evidence that the true mean difference in number of bean that can be placed into a cup in 15 econd (dominant hand nondominant hand) i greater than 0. (We don t have evidence that manual dexterity i better, on average, for the dominant hand. 6) In a tudy of memory recall, eight tudent from a large pychology cla were elected at random and given 10 minute to memorize a lit of 20 nonene word. Each wa aked to lit a many of the word a he or he could remember both 1 hour and 24 hour later. I there evidence to ugget that the mean number of word recalled after 1 hour exceed the mean recall after 24 hour by more than 3 word? Ue.01 ignificance level. Student hr later hour later Difference State: μdiff: the true mean difference in number of word recalled after one hour v. 24 hour (1 hr 24 hr) H0:μ diff = 3 HA: μ diff > 3 Plan: One-ample paired t tet (σ unknown) 1) The problem tated that the ample wa elected at random. 2) We are not told the hape of the population. The boxplot i kewed and there i an outlier. Our ample ize of 9 i not large enough to overcome thee eriou non-normal feature. Proceed with caution. 3) Becaue we ampled without replacement, aume that there are at leat 10(8) = 80 tudent in the population. Alo aume that the difference in the number of word remembered by each tudent wa independent of other tudent. = df = 8 p-value = = Becaue our p-value i larger than our ignificance level of 0.01 we fail to reject H0. There i not ufficient evidence that the true mean number of word recalled after 1 hour exceed the mean recall after 24 hour by more than 3 word.

5 7) Paired t for height momheight N Mean StDev SE Mean Height Momheight Difference % CI for mean difference: (0.639, 1.931) t-tet of mean difference = 0 (v > 0); t-value = 3.95; p-value = a) It ha been hypotheized that college tudent are taller than they were a generation ago and therefore that college women hould be ignificantly taller than their mother. State the null and alternative hypothee to tet thi claim. Be ure to define any parameter you ue. μdiff: the true mean difference in height of college tudent and their mother (current tudent mother) H0: μ diff = 0 HA: μ diff > 0 b) Uing the information in the Minitab output, the tet tatitic i t = Identify the number that were ued to compute the t-tatitic. What formula wa ued to calculate the t-tatitic? x diff = 1.285, n = 93, diff = c) What are the degree of freedom for the tet tatitic? df = 92 d) Write the probability tatement of the hypothei tet. There i almot a 0% chance of getting a ample mean difference in height a extreme at due to ampling variability if the true mean difference in height i 0. Becaue thi i o unlikely, we can reject our null hypothei at any reaonable ignificance level. We have trong evidence that the true mean height of college tudent i greater than the generation before. e) Why i thi a paired t tet? College tudent were paired with their mother. Therefore, the two ample were not independent.