P-values for t-distribution:

Either use the tcdf() command on your calculator (if applicable) or use the following website (using this website during the exam is allowed).

https://homepage.divms.uiowa.edu/~mbognar/applets/t.html

Solution to previous problem: (1) Step 1: Significance level is .05

Step 2: H0: 𝛍1= 𝛍2Step 3: HA: 𝛍1< 𝛍2Step 4: t=(59.4-61.3)/(1.8/ 252+2.4/272 ).5 =-3.245Step 5: t.05= 1.677 so rejection region is (-∞,-1.677]

(2) Step 6: Reject H0

(3) P-value=P(T<-3.245)=0.0011(4) At the 5% significance level, reject the null hypothesis that mean skeletal height of females in

Tribe 1 is the same as that of Tribe 2. There is significant evidence that mean skeletal height of females in Tribe 1 is lower than that of Tribe 2.

(5) Since we rejected H0 , we risk making a type 1 error; the conditional probability of this error, given that the null hypothesis is true, is .05.

Solution to previous problem: (1) Step 1: Significance level is .05

Step 2: H0: 𝛍1= 𝛍2Step 3: HA: 𝛍1≠ 𝛍2Step 4:t=(29.964-24.931)/(5.377/ 112+2.238/132 ).5 =2.899Step 5: t.025= 1.771 so rejection region is (-∞,-2.16]∪[2.16, ∞)

(2) Step 6: Reject H0

(3) P-value=2P(T>2.899)=0.0124(4) At the 5% significance level, reject the null hypothesis that mean IBU of American Brown Ales is

the same as that of English Brown Ales. There is significant evidence that mean IBU of American Brown Ales is lower than that of English Brown Ales.

(5) Since we rejected H0 , we risk making a type 1 error; the conditional probability of this error, given that the null hypothesis is true, is .05.

Solution to previous problem: (1) Step 1: Significance level is .05

Step 2: H0: 𝛍d=0Step 3: HA: 𝛍d<0Step 4: t=(-0.929-0)/(1.699/ 7.5)=-1.473Step 5: t.05= 1.943 so rejection region is (-∞,-1.943]

(2) Step 6: Do Not Reject H0

(3) P-value=P(T<-1.943)=0.0956(4) At the 5% significance level, we do not reject the null hypothesis that mean workplace efficiency

changer per employee, as measured by HES, is the same at the two different office temperature. There is not significant evidence that mean HES is lower at 85 degrees than it is at 80 degrees.

(5) Since we did not reject H0 , we risk making a type 2 error; the conditional probability of this error, represented by a 𝛃-curve, depends on the actual value of 𝛍d.

Solution to previous problem: (1) Step 1: Significance level is .05

Step 2: H0: p=0Step 3: HA: p<0Step 4: z=(98/168-2/3)/ (2/3(1-2/3)/ 168).5=-2.29Step 5: t.05= 1.943 so rejection region is (-∞,-1.645]

(2) Step 6: Reject H0

(3) P-value=P(T<-1.943)=0.011(4) At the 5% significance level, we reject the null hypothesis that mean the true proportion of

Oregon anglers that identify as ‘trout anglers’ is equal to 2/3. There is significant evidence that this proportion is less than 2/3.

(5) Since we rejected H0 , we risk making a type 1 error; the conditional probability of this error, given that the null hypothesis is true, is .05.

Solution to previous problem: (1) Step 1: Significance level is .05

Step 2: H0: p=.5Step 3: HA: p<.5Step 4: z=(207/500-.5)/ (.5(1-.5)/ 500).5=-3.846Step 5: t.05= 1.943 so rejection region is (-∞,-1.645]

(2) Step 6: Reject H0

(3) P-value=P(T<-3.846)=0.0001(4) At the 5% significance level, we reject the null hypothesis that “most American adults have at

least one drink a week” in favor of the alternative hypothesis that true proportion of American adults that have at least one drink a week is less than .5.

(5) Since we rejected H0 , we risk making a type 1 error; the conditional probability of this error, given that the null hypothesis is true, is .05.

Solution to previous problem: (a) See top right. The data does appear to be

scattered around a line(a) Y = -0.9882 + 0.8446 x,

see scatterplot w/ line at lower right(c) We find that r2=0.719

Solution to previous problem: (a) See top right. Although a slight downward

concavity might be perceived, this patternvanishes with the removal of a single datapoint, so the evidence of a nonlinear pattern is weak

(b) y = 3.2187 + 2.004 x, see scatterplot w/ line at lower right

(c) We find that r2=0.8185