Cheat Sheet- Formulas- STATS (2)

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May 19, 2024

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X bar 5.3 mean S 14.6 standard deviation Alpha 0.05 1- the confidence interval N 95 sample size E 2.9741719 confidence.t (alpha,standard deviation,sample size) Lower Upper 2.325828 8.274172 x bar- E x bar + E how to find null and alternate hypothesis ex: a. What is the null & alternate hypothesis should you test? b. Suppose you select a= 0.05. Interpret this value in the words of the problem. *The null is the ONLY alpha, the level of significance, is the probability of observing *Assume the null is TR a Type 1 error. Since alpha= .05, there is a 5% chance that we make a Type 1, that we conclude that the mean expenditure is more than 2400, falsely. c. For a=0.05, specify the rejection region of a large sample test. norm.s.inv (1-alpha) ---> norm.s.inv(1-0.05)= 1.645 Z>1.645 Two-Tailed Hypothesis Test Example PROBLEM A study found that 34% of teenagers text while driving. A recent study by At&T found that 5 selected teens had texted while driving. do the results suggest that the proportion of teens has changed since 2009? use a 95% confidence interval to answer. Step 1: Set up the Hypothesis Step 2: Alpha Step 3: Fin H0: p=0.34 alpha= 0.05 Ha: p=/= 0.34 Example x 515 n 1200 p-hat 0.42917 x divided by n (515/1200) p0 0.34 z0 6.52051 p-value 0 2*norm.s.dist(-abs(6.52051),TRUE) *click format cells, alpha 0.05 1-95% p-value<alpha 0<0.05 REJECT; There's enough evidence to support the that a proportion different from 0.34 of teens hav Review from Chapter 9: E: confidence .norm(alpha.sqrt(p-hat*(1-p-hat)),n E= 0.028 *Another way to determine if you reject is see if the p0 <---- how to find the Things you have to think about what is the parameter? mean expenditure (null) H0: ( mean) 2400 what is the numerical value? MEAN VALUE *ex: 240 ( alternate ) Ha: ( mean )> 2400 what is the alternate hypothesis? upper- tailed tes ( p-hat - p0) - SQRT(p0*(1 - p0)/n)

Lower Upper p-hat - E p-hat + E 0.40116 0.45717 PROBLEM The fun size of a snickers bar is supposed to weight 20 grams. The manufacturer calibrates the machine so that the mean weight is 20.1 grams. The engineer from the manufacturer is concerned so he obtains a sample of 11 random candy bars, weighs them and obtains the data in table 1. should the machine be shut down and calibrated? He decides to conduct the test at the a=0.01 level of significance. Table 1 19.68 20.66 19.56 19.98 20.65 19.61 20.55 20.36 21.01 21.5 19.74 PROBLEM According to the American Community Survey, the mean travel time to work in Colin Count The Department of Transportation reprogrammed all the traffic lights in Collin County in an To determine whether there is evidence that travel time has decreased as a result of the re obtains a random sample of 2500 commuters, record their travel time to work, and finds a s 8.5 minutes. Does this result suggest that travel time decreased at the a=0.05 level of sign H0: mu=27.6 alpha:0.05 x bar 27.3 Ha:mu<27.6 s 8.5 Reject, the n 2500 the averag t0 -1.76471 p-value 0.0338868

10.1 Rejection Region Example: Left Tailed z<z0 norm.s.inv(alpha) Ha: mean> Right Tailed z>z0 norm.s.inv(1-alpha) norm.s.inv( Two Tailed z<-z0 OR z>z0; Ha: mean> z> |z0|: z0= norm.s.inv(1-alpha/2) norm.s.inv( P-Value Tailed Test Right Tailed:P(Z>z0)= 1-norm.s.inv(z0,TRUE) OR norm.s.in Left Tailed: P(Z<z0)=norm.s.inv(z0, TRUE) Two Tailed: P(|Z|>z0)=2*norm.s.inv(-abs(z0),TRUE) OR 2*( thing you can reject* ex: P-Value Tailed Test: RUE* a. Ha: mean>7,z=1.20 P(Z>1.20) b. Ha: mean<7,a=1.20 P(Z<-1.20) c. Ha: mean =/ 7, z=1.20 P(|Z|>1.20 State the Conclusion: p-value < alpha or z is in the reject region: Reject null hypothesis. There's enough evidence to conclud *insert alt hypothesis here* 515 of 1200 randomly who text while driving p-value> alpha or z is not in the rejection region: Fail to reject null hypothesis. There's not enough evidence *insert alt hypothesis here* nd the best stat/p-value/rr Testing H Two Tailed Left Tailed H0:mu=mu0 H0:mu=mu H1:mu=/mu0 H1: mu<m *mu0 is the assumed value of the then number, then decimal places (12)* Step 4. If P value < a, r Step 5. State the concl claim ve texted while driving. To find test statistic (t0 0 lies in between lower and upper* To find x bar = average e upper and lower bound for x bar 00* st

To find s= stdev.s(data To find n= count(data) df= degrees of freedom Step 1: Place all data 19.86 20.66 19.56 19.98 20.65 19.61 20.55 20.36 21.02 21.5 19.74 Step 3: Step 2: Write hypothesis x bar 20.30091 average (d H0: mu= 20.1 alpha: 0.01 s 0.640366 stdev.s(dat Ha: mu =/= 20.1 n 11 count(data df 10 n-1 t0 x (x bar-20.1 p-value 0.322584 t.dist.2t(t0, Step 4: ty, Texas, is 27.6 minutes. Do not Reject, not enough eviden attempt to reduce travel time. that the snickers bars are differen eprogramming, the Department of Transportartion sample mean of 27.3 with a standard deviation of nificance? How to differentiate which to ere's enough evidence to support ge commute time has decreased in collin county

> 4, a=0.05 (0.05)= 1.644854 Right tailed > 4, a=0.10= (0.10)= -1.281552 Left Tailed *abs= absolute value* make the test statistics positive nv(-z0,TRUE) (1-norm.s.inv(abs(z0),TRUE 0.11507 <--- has the be less than alpha in order to REJECT 0.11507 <---- same as above 0) 0.230139 de that to conclude that t.dist.2t(t0,df,TRUE) t.dist.rt(t0, Hypothesis Regarding Mean Right Tailed u0 H0:mu=mu0 mu0 H1:mu>mu0 e poulation mean. reject the null hypothesis. lusion t.dist(t0,df,TRUE) 0): t0=(xbar-mean0)/(s/sqrt(n)) Example: The mean height of American ma ge (data) 44 male US presidents have a mean of 70.8

a) *df is = n-1* Treating the 44 presidents as a random sam ) that the US presidents are taller than the av m Step 1: H0: mu = 69.5 alpha= 0.05 Step 2: Ha: mu > 69.5 x bar s n Step 3: t0=(xbar - mu)/ standard deviation/sqrt(n)) Find t0 3.25588 ---> *rounded*---> 3.256 (test statistic) data) ta) Step 5: a) p-value = 0.0011 p-value > alpha 1)/s/sqrt(n)) Reject, there's not enough evidence to support the claim ,df) that the average presidential height is taller than the average height of an American male. nce to conclude nt average mass of 20.1 use:

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