Mixed Hypothesis Testing Review Include the following a.) type of test b.) null/alternate hypothesis c.) draw a curve labeled with your CV and rejection region d.) Test statistic and calculations if need. e.) conclusion with p-value interpretation f.) A CI to support your decision. Explain how the CI supports your conclusion. 1. The following represents the results of a survey taken at SAC. A random sample of full time students were asked how much they work (Do not work, PT, or FT) and how many years they have been attending SAC. Test if there is an association between education level and employment status at significance level of 0.05 (skip part e). Number of years at SAC

Employment Status

1st year 2nd year 3rd+year

Do not work 72 66.7

43 47.0

20 18.3

Work 88 90.3

65 61.0

22 23.7

a.) 𝜒!  𝑇𝑒𝑠𝑡  𝑜𝑓  𝐼𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑐𝑒  𝑑𝑓 = 3 − 1 2 − 1 = 2

b.) H0: OBS = EXP Number of years at SAC is Independent of employment. Ha: OBS≠EXP There is an association between number of years and employment.

c.) CV = 5.991

d.) 𝜒! = Σ !"#!!"# !

!"#= !"!!!.! !

!!.!+⋯+ !!!!".! !

!".!= 1.03

e.) Since 1.03 is larger than the CV of 5.991, the P-value will not be

unusually small. We fail to reject H0, we do not have enough evidence to prove an association between employment status and number of years at SAC.

f.) NO CI required for this problem

2. The data are of temperature when a 10K was run and of the winning times at those temperatures.

Temp 55 61 49 62 70 73 57 51 Time 32.2 31.5 30.9 31.7 32.9 33.4 32.1 31.4

Residual  standard  error:  0.4496  on  df  =  6  Multiple  R-­‐squared:  0.7435,    Adjusted  R-­‐squared:  0.7008    F-­‐statistic:    17.4  on  1  and  6  DF,    p-­‐value:  0.005874    

Is there evidence of a linear relationship at between temperature and time? Test using 𝛼 = 0.01

a.) Lin regression t-test df = 6

b.) H0: 𝛽 = 0 There is no linear relationship between temp and win time Ha: 𝛽 ≠ 0 There is a linear relationship between temp. and win time. 𝛽 = 𝑠𝑙𝑜𝑝𝑒  𝑏𝑒𝑡𝑤𝑒𝑒𝑛  𝑡𝑒𝑚𝑝  𝑎𝑛𝑑  𝑡𝑖𝑚𝑒

c.) t* = 3.707

d.) Test Stat: t = 4.171

e.) The p-value of 0.00587 is unusually small (statistically significant) at 𝛼 = 0.01. We reject the null hypothesis. Storng evidence there is a linear relationship between temp and 10K winning time.

Coefficients:                               Estimate     Std.  Error     t  value    Pr(>|t|)          (Intercept)       27.04897        1.20064       22.529    5.01e-­‐07  temp                       0.08307           0.01992         4.171       0.00587    

3. A university wanted to see if men or women did better in a chemical engineering course. They found that 23 of the 34 women passed and 60 of the 89 men were successful. Run a complete test to find if there is evidence of a difference between the proportions of women and men who succeed. Use α = .05, TS = 0.161, P-val = 0.987 moe = 0.159

a.) Two sample proportion z – test

b.) H0: p1=p2 There is no difference in success rates between genders. Ha: p1≠p2 There is a difference in success rates between genders.

c.) Z*=1.96

d.) Test Stat: z = 0.161

e.) The p-value of 0.987 is not unusually small. We fail to reject H0, we do not have enough evidence to suggest the proportion of men who were successful is different than the proportion of females.

f.) 𝑝! − 𝑝! = 0.676 − 0.674 = 0.002 CI: ( 0.002 - .159, 0.002 + 0.159) We are 95% confident the true difference in proportions (pwomen-pmen) is between -0.158 and .161). Since 0 is in this confidence interval there could be no difference.

4. Below are the times (in minutes) recorded in a triathlon for one participant? Each mile consists of 5 laps on three different courses. Test the claim that the average times are the same for each mile. Test at the α = .05 significance level. Mile 1 3:15 3:24 3:23 3:22 3:21 Mile 2 3:19 3:22 3:21 3:17 2:19 Mile 3 3:34 3:31 3:29 3:31 3:29

a.) ANOVA b.) H0: 𝜇! = 𝜇! = 𝜇! c.) Skip d.) Test Stat. F = 2.68 e.) The P-value of 0.109 is not unusually small (not statistically significant)

if α = .05, we fail to reject the null hypothesis. There is not enough evidence to suggest a difference of means among the 3 groups.

5. Five engineering students and 5 computer science students are arranged in a matched pair design on the basis of their GPA prior to the last semester of their senior year. Their GPAs for their last semester are as follows: Perform a 95 percent confidence interval to test the claim whether the mean GPA of Engineering majors is less than the mean GPA of CS Majors

a.) Matched pair t-test df = 4

b.) H0: 𝜇!"# = 0 difference = Engineering GPA – CS GPA HA: 𝜇!"# < 0

c.) Critical value: t* = 2.776 𝑥!"# = 0.524            𝑠!"# = 1.465

𝑥 ± 𝑡∗𝑠𝑛= 0.524 ± 2.776

1.4655

= (−1.30, 2.34)

d.) I am 95% confident the true mean difference (difference = Engineering

GPA – CS GPA) is between -1.30 (Engineering is lower than CS) to 2.34 (Engineering’s GPAs are higher than CS). A difference of 0 is in the confidence interval, evidence is inconclusive. We Fail to reject the null hypothesis. We do not have enough evidence to suggest on average Engineering has a lower GPAs than CS majors.

Pair Number Engineering Major CS Major E - CS 1 3.89 3.75 0.14 2 3.61 3.87 -0.26 3 2.80 3.02 -0.22 4 2.75 2.92 -0.17 5 4.41 1.28 3.13

6. A survey of 4000 people in the US finds that 2856 of them believe that daily weather reports are totally useless because meteorology is not really a science. Given this data perform a 99 percent hypothesis test to see if more than half of the people in the US believe that weather reports are useless.

a.) 1 sample proportion z – interval

b.) H0: p = 0.5 p = prop. of weather reports “useless” Ha: p > 0.05

c.) CV z*=2.576 𝑝 = !"#$!"""

= 0.714

𝑝 ± 𝑧∗𝑝𝑞𝑛 =

0.714± 2.5760.714 0.286

4000 =

(0.696, 0.732)

d.) I am 99% confident that the true proportion of people in the US who

believe daily weather reports are useless is between 69.6% and 73.2%. Both bounds of the interval are greater than 50%. We reject the null hypothesis. There is strong evidence that more than half of people in the US believe the weather report is useless.